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MATHEMATICAL TEXT- BOOKS 
By G. A. WENTWORTH, A.M. 



Mental Arithmetic. 

Elementary Arithmetic. 

Practical Arithmetic. 

Primary Arithmetic. 

Grammar School Arithmetic. 

High School Arithmetic. 

High School Arithmetic (Abridged). 

First Steps in Algebra. 

School Algebra. 

College Algebra. 

Elements of Algebra. 

Complete Algebra. 

Shorter Course in Algebra. 

Higher Algebra. 

New Plane Geometry. 

New Plane and Solid Geometry. 

Syllabus of Geometry. 

Geometrical Exercises. 

Plane and Solid Geometry and Plane Trigonometry. 

New Plane Trigonometry. 

New Plane Trigonometry, with Tables. 

New Plane and Spherical Trigonometry. 

New Plane and Spherical Trig., with Tables. 

New Plane and Spherical Trig., Surv., and Nav. 

New Plane Trig, and Surv., with Tables. 

New Plane and Spherical Trig., Surv., with Tables. 

Analytic Geometry. 



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A 



TEXT-BOOK 



OP 



GEOMETEY. 



REVISED EDITION. 



BY 



G. A. WENTWORTH, A.M., 

AUTHOK OF A SB&IBS OF TEXT-BOOKS IN MATHEMATICS. 



BOSTON, U.S.A.: 
PUBLISHED BY GINN & COMPANY. 

1838— 



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Entered, according to Act of Congreu« in the year 1888, by 

a. A. WENTWORTH, 
in the Office of the Librarian of Congress, at Washington. 



All Rights Rbsbryed. 



Ttpoorapht bt J. 8. Gushing & Co., Boston, U.SA^ 



Pbssswobk bt Ginn & Co., Boston, U.B.A. 



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PREFACE. 



TV ^OST persons do not possess, and do not easily acquire, the power 
-*-^-*- of abstraction requisite for apprehending geometrical concep- 
tions, and for keeping in mind the successive steps of a continuous 
argument. Hence, with a very large proportion of beginners in Geom- 
etry, it depends mainly upon the form in which the subject is pre- 
sented whether they pursue the study with indifference, not to say 
aversion, or with increasing interest and pleasure. 

In compiling the present treatise, the author has kept this fact con- 
stantly in view. All unnecessary discussions and scholia have been 
avoided; and such methods have been adopted as experience and 
attentive observation, combined with repeated trials, have shown to be 
most readily comprehended. No attempt has been made to render 
more intelligible the simple notions of position, magnitude, and direc- 
tion, which every child derives from observation ; but it is believed 
that these notions have been limited and defined with mathematical 
precision. 

A few symbols, which stand for words and not for operations, have 
been used, but these are of so great utility in giving %iyU and -ptr- 
spicuity to the demonstrations that no apology seems necessary for 
their introduction. 

Great pains have been taken to make the page attractive. The 
figures are large and distinct, and are placed in the middle of the 
page, so that they fall directly under the eye in immediate connec- 
tion with the corresponding text. The given lines of the figures are 
full lines, the lines employed as aids in the demonstrations are short- 
dotted, and the resuUtng lines are long-dotted. 



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IV PREFACE. 

In each proposition a concise statement of what is given is printed 
in one kind of type, of what is required in another, and the demon- 
Btration in still another. The reason for each step is indicated in 
small type between that step and the one following, thus preventing 
the necessity of interrupting the process of the argument by referring 
to a previous section. The number of the section, however, on which 
the reason depends is placed at the side of the page. The constituent 
parts of the propositions are carefully marked. Moreover, each distinct 
assertion in the demonstrationit and each particular direction in the 
construction of the figures, begins a new line; and in no case is ii 
necessary to turn the page in reading a demonstration. 

This arrangement presents obvious advantages. The pupil perceives 
at once what is given and what is required, readily refers to the figure 
at every step, becomes perfectly familiar with the language of Geom- 
etry, acquires facility in simple and accurate expression, rapidly learns 
to reason^ and lays a foundation for completely establbhing the 
science. 

Original exercises have been given, not so difficult as to discourage 
the beginner, but well adapted to afford an effectual test of the degree 
in which he is mastering the subjects of his reading. Some of these 
exercises have been placed in the early part of the work in order 
that the student may discover, at the outset, that to commit to mem- 
ory a number of theorems and to reproduce them in an examination 
is a useless and pernicious labor ; but to learn their uses and appli- 
cations, and to acquire a readiness in exemplifying their utility is to 
derive the fall benefit of that mathematical training which looks not 
80 much to the attainment of information as to the discipline of the 
mental faculties. 

G. A. WENTWORTH. 
Exeter, N.H. 
1878. 



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PREFACE. 



TO THE TEACHER. 

Wheit the pupil is reading each Book for the first time, it will be 
well to let him write his proofs on the blackboard in his own lan- 
guage ; care being taken that his language be the simplest possible, 
that the arrangement of work be vertical (without side work), and 
that the figures be accurately constructed. 

This method will fnrnish a valuable exercise as a language lesson, 
will cultivate the habit of neat and orderly arrangement of work, 
and will allow a brief interval for deliberating on each step. 

After a Book has been read in this way, the pupil should review 
the Book, and should be required to draw the figures free-hand. He 
should state and prove the propositions orally, using a pointer to 
indicate on the figure every line and angle named. He should be 
encouraged, in reviewing each Book, to do the original exercises ; to 
state the converse of propositions ; to determine from the statement, 
if possible, whether the converse be true or false, and if the converse 
be true to demonstrate it ; and also to give well-considered answers 
to questions which may b^ asked him on many propositions. 

The Teacher is strongly advised to illustrate, geometrically and 
arithmetically, the principles of limits. Thus a rectangle with a con- 
stant base 6, and a variable altitude x, will afford an obvious illus- 
tration of the axiomatic truth that the product of a constant and a 
variable is also a variable ; and that the limit of the product of a 
constant and a variable is the product of the constant by the limit 
of the variable. If x increases and approaches the altitude a as a 
limit, the area of the rectangle increases and approaches the area of 
the rectangle a6 as a limit; if, however, x decreases and approaches 
zero as a limit, the area of the rectangle decreases and approaches 
zero for a limit. An arithmetical illustration of this truth may be 
given by multiplying a constant into the approximate values of any 
repetend. If, for example, we take the constant 60 and the repetend 
0.3333, etc., the approximate values of the repetend will be ^, •^, 



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VI PREFACE. 

^ftAr» ^Aftflr* ®*c-t wid ^^68© values multiplied by 60 give the serieB 
18, 19.8, 19.98, 19.998. etc., which evidently approaches 20 as a limit; 
but the product of 60 into i (the limit of the repetend 0.333, etc.) is 
also 20. 

Again, if we multiply 60 into the different values of the decreasing 
seriee A, y^, y^^, ji^jfjr* «*«•. which approaches zero as a limit, we 
shall get the decreasing seriee 2, }, ^, y^, etc.; and this series evi- 
dently approaches zero as a limit 

In this way the pupil may easily be led to a complete compre- 
hension of the subject of limits. 

The Teacher is likewise advised to give frequent written examina- 
tions. These should not be too difficult, and sufficient time should be 
allowed for accurately constructing the figures, for choosing the best 
language, and for determining the best arrangement 

The time necessary for the reading of examination-books will be 

diminished by more than one-half, if the use of the symbols employed 

in this book be allowed. 

Q. A. W. 
Exeter, N.H. 

1879. 



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PEEFACB. Vll 



NOTE TO REVISED EDITION. 

The first edition of this Geometry was issued about nine years ago. 
The book was received with such general favor that it has been neces- 
sary to print very large editions every year since, so that the plates 
are practically worn out. Taking advantage of the necessity for new 
plates, the author has re-written the whole work ; but has retained 
all the distinguishing characteristics of the former edition. A few 
changes in the order of the subject-matter have been made, some of 
the demonstrations have been given in a more concise and simple 
form than before, and the treatment of Limits and of Loci has been 
made as easy of comprehension as possible. 

More than seven hundred exercises have been introduced into this 
edition. Tliese exercises consist of theorems, loci, problems of con- 
struction, and problems of computation, carefully graded and specially 
adapted to beginners. No geometry can now receive favor unless it 
provides exercises for independent investigation, which must be of such 
a kind as to interest the student as soon as he becomes acquainted 
with the methods and the spirit of geometrical reasoning. The author 
has observed with the greatest satisfaction the rapid growth of the 
demand for original exercises, and he invites particular attention to 
the systematic and progressive series of exercises in this edition. 

The part on Sdltf Geometry has been treated with much greater 
freedom than before, and the formal statement of the reasons for the 
separate steps has )i^kn in general omitted, for the purpose of giving a 
more elegant form t6 the demonstrations. 

A brief treatise dn Conic Sections (Book IX) has been prepared, 
and is issued in pamphlet form, at a very low price. It will also be 
bound with the Geometry if that arrangement is found to be gen- 
erally desired. 



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viii PREFACE. 

The author takes this opportunity to express his grateful appre- 
ciation of the generous reception given to the Geometry heretofore by 
the great body of teachers throughout the country, and he confidently 
anticipates the same generous judgment of his efforts to bring the work 
up to the standard required by the great advance of late in the sci- 
ence and method of teaching. 

The author is indebted to many correspondents for valuable sug- 
gestions ; and a special acknowledgment is due, for criticisms and 
careful reading of proofs, to Messrs. C. H. Judson, of Greenville, S.C. ; 
Samuel Hart, of Hartford, Conn. ; J. M. Taylor, of Hamilton, N.Y. ; 
W. Le Conte Stevens, of Brooklyn, N.Y. ; E. R. Offutt, of St. Louis, 
Mo.; J. L. Patterson, of Lawrenceville, N. J.; G. A. Hill, of Cam- 
bridge, Mass.; T. M. Blakslee, of Des Moines, la.; G. W. Sawin, of Cam- 
bridge, Mass. ; Ira M. De Long, of Boulder, Col. ; and W. J. Lloyd, of 
New York, N.Y. 

Corrections or suggestions will be thankfully received. 

G. A. WENTWORTH. 

EXETEB, N.H., 

1888. 



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OOI^TEl^TS. 



GEOMETRY. 

PAOS 

Defikitiovs 1 

Stbaight Lines 6 

Plane Angles 7 

Magnitude of Angles 9 

Angular Units 10 

Method of Supeeposition 11 

Symmetet .13 

Mathematical Teems 14 

Postulates 15 

Axioms . . .16 

Symbols 16 



PLANE GEOMETRY. 

BOOK I. The Straight Line. 

The Straight Line 17 

Parallel Lines 22 

Perpendicular and Oblique Lines 33 

Triangles 40 

Quadrilaterals 56 

Polygons in General 66 

E;cBRCwa» 72 

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X CONTENTS. 

BOOK II. The Circle. 

PAGS 

Definitions 75 

Arcs and Chobds 77 

Tangents 89 

Measubement . 92 

Theoby op Limits 94 

Measube of Angles 98 

Pboblems op Constbuction 106 

Ezebcises 126 

BOOK III. Proportional Lines and Similar Polygons. 

Theoby op Pbopobtion 131 

Pbopobtional Lines 138 

SiifiLAB Tbiangles 145 

SiMiLAB Polygons 153 

Numebical Propebties of Figubes 156 

Pboblems op Constbuction 167 

Pboblems op Computation 173 

Exercises 175 

BOOK IV. Areas of Polygons. 

Abeas op Polygons 180 

compabison of polygons 188 

Pboblems of Constbuction 192 

Pboblems of Computation 204 

Ezebcises , 205 

BOOK V. Regular Polygons and Circles. 

Regulab Polygons and Cibgles 209 

Pboblems of Constbuction 222 

Maxima and Minima 230 

Exebcises 237 

misoellaneous exebcises 240 



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GEOMETRY. 



DEFINITIONS. 

1. If a block of wood or stone be cut in the shape repr^ 
Beuted in Fig. 1, it will have six flat faces. 

Each face of the block is called 
a surface; and if these faces are made 
smooth by polishing, so that, when 
a straight-edge is applied to any one 
of them, the straight edge in every 
part will touch the surface, the faces 
are called plane surfaces^ or planes, 

2. The edge in which any two of these surfaces meet is 
called a line, 

8. The comer at which any three of these lines meet is 
called a point, 

4. For computing its volume, the block is measured in three 
principal directions : 

From left to right, Aio B, 
From front to back, A to C, 
From bottom to top, A to D, 

These three measurements are called the dimensions of the 
block, and are named lengthy breadth (or width), thickness 
(heiffht OT depth). 



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2.'. •:.;•/: ••^•': : : .-.aEOMETRY. 

A solidj therrforef has three dimensionSy length, breadth, and 
thickness. 

6. The surface of a solid is no part of the solid. It is 
simply the boundary or limit of the solid. A surface^ there- 
fore, has only two dimensions, length and breadth. So that, 
if any number of flat surfaces be put together, they will 
coincide and form one surface. 

6. A line is no part of a surface. It is simply a boundary 
or limit of the surface. A line, therefore, has only one dimen- 
sion, length. So that, if any number of straight lines be put 
together, they will coincide and form one line. 

7. A point is no part of a line. It is simply the limit of 
the line. A point, therefore, has no dimension, biU denotes 
position simply. So that, if any number of points be put 
together, they will coincide and form a single point. 

8. A solid, in common language, is a limited portion of 
space filled mith matter ; but in Geometry we have nothing 
to do with the matter of which a body is composed ; we study 
simply its shape and size; that is, we regard a solid as a 
limited portion of space which may be -occupied by a physical 
body, or marked out in some other way. Hence, 

A geometrical solid is a limited portion of space, 

9. It must be distinctly understood at the outset that the 
points, lines, surfaces, and solids of Geometry are purely 
ideal, though they can be represented to the eye in only a 
material way. Lines, for example, drawn on paper or on the 
blackboard, will have some width and some thickness, and 
will so far fail of being true lines; yet, when they are used to 
help the mind in reasoning, it is assumed that they represent 
perfect lines, without breadth and without thickness. 



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DEFINITIONS. 3 

lOt A point is r^eaented to the eye by a fine dot, and 
named by a letter, as A (Fig. 2) ; a line is named by two 
letters, placed one at each end, 
as BF\ a surface is represented 
and named by the lines which 
bound it, as BODF\ a solid is 
represented by the faces which 
bound it. Fi<*- 2. 

11. By supposing a solid to diminish gradually until it 
vanishes we may consider the vanishing point, a 'pcird in 
space, independent of a line, having position but no extent 

12. If a point moves continuously in space, its path is a 
line. This line may be supposed to be of unlimited extent, 
and may be considered independent of the idea of a surface. 

13. A surface may be conceived as generated by a line 
moving in space, and as of unlimited extent. A surface can 
then be considered independent of the idea of a solid. 

14. A solid may be conceived as generated by a surface in 
motion. 

Thus, in the diagram, let the up- ^ ^ 

right surface AJBCD move to the 
right to the position EFGH. The 

points A, B, (7, and D will generate ^ , . 

the lines AE, BF, CG, and DH, ^ 
respectively. The lines AB, BC, -, « 

CDy and AD will generate the sur- 
faces AF, BQ, OH, and AH, respectively. The surface 
ABCD will generate the solid AO, 

15. Oeomdry is the science which treats of position, form, 
and Tnagnitude. 

16. Points, lines, surfaces, and solids, with their relations, 
constitute the subject-matter of Geometry. 



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4 GEOMETRY. 

17. A straight line, or right line^ is a line which has the 

same direction throughout its ^ ^ 

whole extent, as the line AB. 

18. A curved line is a line 
no part of which is straight, 
as the line CD. 

19. A broken line is a series 
of different successive straight 
lines, as the line ER 

20. A mixed line is a line composed of straight and curved 
lines, as the line OH. 

A straight line is often called simply a line, and a curved 
line, a curve. 

21. A plane m/rface, or a plane, is a surface in which, if 
any two points be taken, the straight line joining these points 
will lie wholly in the surface. 

22. A curved surface is a surface no part of which is plane. 

23. Figure or form depends upon the relative position of 
points. Thus, the figure or form of a line (straight or curved) 
depends upon the relative position of the points in that line ; 
the figure or form of a surface depends upon the relative 
position of the points in that surface. 

24. With reference to form or shape, lines, surfaces, and 
solids are called figures. 

With reference to exteifU, lines, surfaces, and solids are 
called magnitudes. 

25. A plane figure is a figure all points of which are in the 
same plane. 

26. Plane figures formed by straight lines are called rec- 
tilinear figured: ; those formed by curved lines are called 
curvilinear figures ; and those formed by straight and curved 
lines are called mixtilinear figures. 



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DEFINITIONS. 5 

27. Figures which have the same shape are called similar 
figures. Figures which have the same size are called equiva- 
lent figures. Figures which have the same shape and size are 
called eqiuil or congruent figures. 

28. Geometry is divided into two parts, Plane Geometry 
and Solid Geometry. Plane Geometry treats of figures all 
points of which are in the same plane. Solid Geometry 
treats of figures all points of which are not in the same plane. 

Straight Lines. 

29. Through a point an indefinite number of straight lines 
may be drawn. These lines will have difierent directions. 

30. If the direction of a straight line and a point in the 
line are known, the position of the line is known ; in other 
words, a straight line is determined if its direction and one of 
its points are known. Hence, 

All straight lines which pass through the same point in the 
same direction coincide, and form but one line* 

31. Between two points one, and only one, straight line 
can be drawn ; in other words, a straight line is determined 
if two of the points are known. Hence, 

Two straight lines which have two points in common coincide 
throughout their whole extent, and form hut one line. 

32. Two straight lines can intersect (cut each other) in only 
one point ; for if they had two points common, they would 
coincide and not intersect. 

33. Of all lines joining two points the shortest is the straight 
line, and the length of the straight line is called the distance 
between the two points. 



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6 GEOMETRY. 

84. A straight line determined by two points is considered 
as prolonged indefinitely both ways. Such a line is called an 
indefinite straight line, 

85. Often only the part of the line between two fixed points 
is considered. This part is then called a segment of the line. 

For brevity, we say "the line AB'* to designate a segment 
of a line limited by the points A and B. 

86. Sometimes, also, a line is considered as proceeding from 
a fixed point and extending in only one direction. This fixed 
point is then called the origin of the line. 

87. If any point C be taken in a given straight line AB^ the 
two parts CA and CB aro 

said to have opposite direc- A ^ B 

tions from the point C. Fio. 5. 

88. Every straight line, as AB, may be considered as hav- 
ing opposite directions, namely, from A towards -B, which is 
expressed by saying " line AB''] and from B towards A, which 
is expressed by saying **line BA.'* 

89. If the magnitude of a given line is changed, it becomes 
longer or shorter. 

Thus (Fig. 5), by prolonging AC to B we add CB to AC, 
and AB = AC+ CB. By diminishing AB to (7, we subtract 
CB from AB, and AC= AB - CB. 

If a given line increases so that it is prolonged by its own 
magnitude several times ^^ ^^ ^ q j) e 

succession, the line is muUi' h ' • ' ^ 

plied, and the resulting line 

is called a multiple of the given line. Thus (Fig. 6), if 

AB = BC== CD^DE, then AC=-2AB, AI) = S AB, and 

AE=^AB. Aho,AB=iAC,AB^iAI),a.ndAB=iAK 

Hence, 



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DEFINITIONS. 



Lines of given length may he added and suitracted; 
may also be multiplied and divided by a nvt/mher. 




Plane Angles. 

40. The opening between two straight lines which meet is 
called a phmje angle. The two lines are called the sides, and 
the point of meeting, the vertex, of the angle. 



41. If there is but one angle at a 
given vertex, it is designated by a cap- 
ital letter placed at the vertex, and is 
read by simply naming the letter ; as, 
angle A (Fig. 7). 

But when two or more angles have 
the same vertex, each angle is desig- 
nated by three letters, as shown in 
Fig. 8, and is read by naming the 
three letters, the one at the vertex be- 
tween the others. Thus, the angle 
DAQ means the angle formed by the 
sides AD and AC, 

It is often convenient to designate 
an angle by placing a small italic let- 
ter between the sides and near the 
vertex, as in Fig. 9. 

42. Two angles are eqvxxl if they 
can be made to coincide. 





Fio. 9. 



43. If the line AD (Fig. 8) is drawn so as to divide the 
angle BAQ into two equal parts, BAD and CAD^ AD is 
called the bisector of the angle BAC. In general, a line that 
divides a geometrical magnitude into two equal parts is called 
a bisector of it. 



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8 



OEOMETRT. 



41 Two angles are called ad' 
jacent when they have the same 
vertex and a common side be- 
tween them ; as, the angles BOD 
and AOD (Fig. 10). 

45. When one straight line 
stands upon another straight line 
and makes the adjacent angles 
eqicalf each of these angles is 
called a right angle. Thus, the 
equal angles DCA and DCB 
(Fig. 11) are each a right angle. 




Fio. 10. 



C 
Fio. 11. 



46. When the sides of an an- 
gle extend in opposite directions, 

so as to be in the same straight line, the angle is called a 
straight angle. Thus, the angle formed at (Fig. 11) with 
its sides CA and CB extending in opposite directions from 0, 
is a straight angle. Hence a right angle may be defined as 
half a straight angle, 

47. A perpendicular to a straight line is a straight line that 
makes a right angle with it. Thus, if the angle DCA (Fig. 11) 
is a right angle, DC is perpendicular to AB, and AB is per- 
pendicular to DC. 

48. The point (as 0, Fig. 11) where a perpendicular meets 
another line is called ihe/oot of the perpendicular. 

49. Every angle less than a right an- 
gle is called an acute angle; as, angle A. 

Fig 12. 

50. Every angle greater than a right 

angle and less than a straight angle is called an obtuse angle; 
afl, angle (7 (Fig. 13). 




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DEFINITIONS. 



61. Every angle greater than a straight angle and less 
than two straight angles is called a reflex angle; as, angle 
(Fig. 14), 




Fio. 13. 



Fio. 14. 



52. Acute, obtuse, and reflex angles, in distinction from 
right and straight angles, are called oblique angles ; and inter- 
secting lines that are not perpendicular to each other are 
called oblique lines, 

53. When two angles have the same vertex, and the sides 
of the one are prolongations of 
the sides of the other, they are 
called vertical angles. Thus, a 
and b (Fig. 15) are vertical an- 
gles. 




Fig. 15. 



54* Two angles are called 
compleinentaiy when their sum 
is equal to a right angle ; and each is called the complement 
of the other; as, angles DOB and DOC {Y'lg. 10). 

55. Two angles are called supplemeniary when their sum is 
equal to a straight angle ; and each is called the supplement 
of the other; as, angles DOB and DO A (Fig. 10). 



Magnitude of Angles. 

66. The size of an angle depends upon the extent of opening 
of its sides, and not upon their length. Suppose the straight 



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10 GEOMETRY. 

line OC to move in the plane of the paper from coincidence 

with OA, about the point as a pivot, to the position 0C\ 

then the line OC describes or generates 

the angle AOO, and the magnitude of the 

angle AOC depends upon the amount 

of rotation of the line from the position 

OA to the position OC. 

If the rotating line moves from the 
position OA to the position OBy perpen- 
dicular to OA, it generates the right 
angle AOB ; if it moves to the position 
OD, it generates the obtuse angle AOD ; if it moves to the posi- 
tion 0A\ it generates the straight angle AOA* ; if it moves to 
the position 0B\ it generates the reflex angle AOB\ indicated 
by the dotted line ; and if it continues its rotation to the posi- 
tion OA, whence it started, it generates two straight angles. 

Hence the whole angular magnitude about a point in a 
plane is equal to two straight angles, or four right angles; and 
the angular magnitude about a point on one side of a straight 
line drawn through that point is equal to one straight angle, 
or two right angles. 

Angles are magnitudes that can be added and subtracted ; 
they may also be multiplied and divided by a number. 

Angular Units. 

57. If we suppose OC (Fig. 17) to 
turn about from a position coinci- 
dent with OA until it makes a com- 
plete revolution and comes again into 
coincidence with OA, it will describe 
the whole angular magnitude about 
the point 0, while its end point C 
will describe a curve called a circum- 
ference. 




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DEFINITIONS. 11 

58. By adopting a suitable unit of angles we are able to 
express the magnitudes of angles in numbers. 

If we suppose OC (Fig. 17) to turn about from coinci- 
dence with OA until it makes one three hundred and sixtieth 
of a revolution, it generates an angle at 0, which is taken 
as the unit for measuring angles. This unit is called a 
degree. 

The degree is subdivided into sixty equal parts called 
minutes, and the minute into sixty equal parts, called seconds. 

Degrees, minutes, and seconds are denoted by symbols. 
Thus, 5 degrees 13 minutes 12 seconds is written, 5** 13' 12". 

A right angle is generated when OC has made one-fourth 
of a revolution and is an angle of 90**; a straight angle is 
generated when OC has made one-half of a revolution and 
is an angle of 180** ; and the whole angular magnitude about 
is generated when OC has made a complete revolution, and 
contains 360**. 

The natural angular unit is one complete revolution. But 
the adoption of this unit would require us to express the 
values of all angles by fractions. The advantage of using the 
degree as the unit consists in its convenient size, and in the fact 
that 360 is divisible by so many diflferent integral numbers. 



Method op Superposition. 

59. The test of the equality of two geometrical magnitudes 
is that they coincide throughout their whole extent. 

Thus, two straight lines are equal, if they can be so placed 
that the points at their extremities coincide. Two angles are 
equal, if they can be so placed that they coincide. 

In applying this test of equality, we assume that a line may 
be moved from one place to another without altering its length; 
that an angle may be taken up, turned over, and put down, 
without altering the difference in direction of its sides. 



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12 




Fio. 18. 

This method enables us to compare magnitudes of the same 
kind. Suppose we have two angles, ABC and DEF. Let 
the side ED be placed on the side BA, so that the vertex E 
shall fall on B ; then, if the side EF falls on BC, the angle 
DEF equals the angle ABC\ if the side EF falls between 
^(7 and BA in the direction BO, the angle DEF is less than 
ABC; but if the side -Ei^ falls in the direction BS", the angle 
DEF ia greater than ABC, 

This method enables us to add magnitudes of the same kind. 

Thus, if we have two straight lines BC_ 

AB and CD, by placing the point ^ ^ 

C on B, and keeping CD in the ^ q 

same direction with AB, we shall Fio. 19. 

have one continuous straight line AD equal to the sum of 

the lines AB and CD. 




jh:. 



r 



Fig. 20. 



B 

Fia. 21. 



ftin : if we have the angles ABC and DEF, and place 
the vertex E on B and the side ED in the direction of BC, the 
angle i>^i^ will take the position CB£[, and the angles DEF 
and ABC will together equal the angle ABH. 

If the vertex E is placed on B, and the side ED on 5.4, the 
angle DEF will take the position ABF, and the angle FBC 
will be the diflference between the angles -4-BCand DEF, 



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DEFINITIONS. 



13 



Xi- 



Fio. 23. 



Symmetry. 

60. Two points are said to be symmetrical with respect to a 
third point, called the centre of sym- ^ 

metry^ if this third point bisects the P' 1 p 

straight line which joins them. Thus, ^^®- ^• 

F and P' are symmetrical with respect to (7 as a centre, if C 

bisects the straight line PP^. 

6L Two points are said to be sym- 
metrical with respect to a straight 
line, called the axis of symmetry, if 
this straight line bisects at right 
angles the straight line which joins 
them. Thus, P and P' are symmet- 
rical with respect to JTJf as an axis, 
if XJf bisects PP^ at right angles. 

62. Two figures are said to be sym- 
metrical with respect to a centre or 
an axis if every point of one has a 
corresponding symmetrical point in 
the other. Thus, if every point in 
the figure A^B^C* has a symmetrical 
point in ABC, with respect to i> as 
a centre, the figure A^B^C is sym- 
metrical to ABO with respect to D 
as a centre. 

63. If every point in the figure 
A^B^O^ has a symmetrical point in 
ABG, with respect to JTJT' as an 
axis, the figure A}B^C^ is symmetri- 
cal to ABO with respect to XX^ as 
an axis. 



A^—^. 



d:*'"' 




Fig. 24. 






Fro. 25. 



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14 



GEOMETRY. 



64. A figure is sTmmetrical with re- 
spect to a point, if the point bisects 
every straight line drawn through it 
and terminated by the boundary of the 
figure. 

66. A plane figure is symmetrical with 
respect to a straight line, if the line 
divides it into two parts, which are sym- 
metrical with respect to this straight 
line. 

Mathematical Terms. 




Fxo. 27. 



66. A 'proof or demonstration is a course of reasoning by 
which the truth or falsity of any statement is logically 
established. 

67. A theorem is a statement to be proved. 

68. A theorem consists of two parts: the hypothesis, or 
that which is assumed ; and the conclusion, or that which is 
asserted to follow from the hypothesis. 

69. An axiom is a statement the truth of which is admitted 
without proof. 

70. A construction is a graphical representation of a geo- 
metrical figure. 

71. A problem is a question to be solved. 

72. The solution of a problem consists of four parts : 

(1) The analysis, or course of thought by which the con- 
struction of the required figure is discovered ; 

(2) The construction of the figure with the aid of ruler and 



(3) The proof that the figure satisfies all the given condi- 
tions; 



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DEFINITIONS. 15 

(4) The discussion of the limitations, which often exist, 
within which the solution is possible. 

73. A postulate is a construction admitted to be possible. 

74. A preposition is a general term for either a theorem or 
a problem. 

75. A corollary is a truth easily deduced from the propo- 
sition to which it is attached. 

76. A scholium is a remark upon some particular feature 
of a proposition. 

77. The converse of a theorem is formed by interchanging 
its hypothesis and conclusion. Thus, 

If A is equal to B, C is equal to D. (Direct.) 
If C is equal to D, A is equal to B. (Converse.) 

78. The opposite of a proposition is formed by stating the 
negative of its hypothesis and its conclusion. Thus, 

If A is equal to B, is equal to D. (Direct.) 

If A is not equal to B, C is not equal to D. (Opposite.) 

79. The converse of a truth is not necessarily true. Thus, 
Every horse is a quadruped is a true proposition, but the 
converse. Every quadruped is a horse, is not true. 

60, If a direct proposition and its converse are true, the 
opposite proposition is true; and if a direct proposition and its 
opposite are true, the converse proposition is true. 

81. Postulates. 

Let it be granted — 

1. That a straight line can be drawn from any one point 
to any other point. 

2. That a straight line can be produced to any distance, 
or can be terminated at any point. 

8. That a circumference may be described about any point 
as a centre with a radius of given length. 



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16 



GEOMETRY. 



82. 



Axioms. 



1. Things which are equal to the same thing are equal to 
each other. 

2. If equals are added to equals the sums are equal. 

3. If equals are taken from equals the remainders are equal. 

4. If equals are added to unequals the sums are unequal, 
and the greater sum is obtained from the greater magnitude. 

5. If equals are taken from unequals the remainders are 
unequal, and the greater remainder is obtained from the 
greater magnitude. 

6. Things which are double the same thing, or equal 
things, are equal to each other. 

7. Things which are halves of the same thing, or of equal 
things, are equal to each other. 

8. The whole is greater than any of its parts. 

9. The whole is equal to all its parts taken together. 



83. 



Symbols and Abbreviations. 



+ increased by. circle. 

— diminished by. Def. . . . 

X multiplied by. Ax. . . . 

-7- divided by. Hyp. . . 

= is (or are) equal to. Cor. . . . 

s:= is (or are) equivalent to. Adj. . . . 

> is (or are) greater than. Iden. . . 

-< is (or are) less than. Cons. . . 

.*. therefore. Sup. . . . 

Z. angle. Siip.-adj. 

jangles. Ext. -int. 

± perpendicular. Alt.-int. 

Jl perpendiculars. Ex. . . . 

II parallel. rt . . . . 

Il» parallels. st 

A triangle. <i.E.D. . . 

^ triangles. 

O parallelogram. a.E.r. . . 
m parallelograms. 



® circles, 
definition, 
axiom, 
hypothesis, 
corollary, 
adjacent, 
identical, 
construction, 
supplementary, 
supplementary, 
exterior-interior, 
alternate-interior, 
exercise, 
right, 
straight, 
quod erat demonstrandum, 

which was to be proved. 
quod erat faciendum, 

which was to be done. 



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PLAICE GEOMETRY. 



BOOK I. 
THE STRAIGHT LIITB. 



Proposition I. Theorem. 
84i All straight angles are equal. 

A e B 

D ^ F 



Let ZBCA and Z FED be any two straight angles. 

To prove Z BOA = Z FED. 

Proof. Apply the Z BCA to the Z FED, so that the vertex 
C shall fall on the vertex E, and the side CB on the side EF. 

Then CA will coincide with ED, 
(5eeauM BCA av^d FED art straight lines and have two points common). 

Therefore the Z BCA is equal to the Z FED. § 59 

a c. D. 

85. CoR. 1. All right angles are equal Ax. 7. 

86. CoR. 2. The angular units have constant values. 

87. CoR. 3. The complements ofegual angles are equal. Ax. 3. 

88. CoR. 4. The supplements of equal angles are equal. Ax. 3. 

89. CoR. 6. At a given point in a given straight line one 
perpendicular, and only one, can be erected. 

HiKT. Consider the given point as the vertex of a straight angle, and 
draw the bisector of the angle. 



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18 PLANE GEOMETKY, — BOOK I. 



Pbopositioh II. Theobem. 

90. If two O/djacent angles have their exterior sides 
in a straight line, these angles are supplements of 
each other* 




Let the exterior sides OA and OB of the Mtfncent 
AAOD »Bd BOD be in the straii^ht line AB. 

To prove A AOD and BOD supplementary. 

Proof. AOB is a straight line. Hjrp. 

.-. the Z AOB is a st. A. § 46 

But the Z AOD + Z BOD = the st. Z AOB. Ax, 9 

/. the A AOD and BOD are supplementary. § 55 

aE.0. 

01. Scholium. Adjacent angles that are supplements of 
each other are called supplementary-adjacent angles, 

02. Gob. Since the angular magnitude about a point is 
neither increased nor diminished by the number of lines which 
railiate from the point, it follows that, 

The sum of all the angles abovi a point in a plane is eqttal 
to two straight angles ^ or four right angles, 
. ITie sum of all the angles about a point on the same side of a 
straight line passing through the poirU ia equal to a straighi 
angle, or two right angles. 



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THE STBAIGHT LIKB. 19 



Proposition III. Theoebm. 

98. Conversely : If two adjacent angles are swpple^ 
menta of each other, their exterior sides lie in the 
same straight line. 



AC B F 

Let the adjacent A OCA + 0CB = 2rt.A. 

To prove AC and CB in the same straight line. 

Proof. Suppose CF to be in the same line with .^(7. § 81 

Then /.0CA + Z.0CF=2Ti.A, §90 

{being sup.-adj. A), 

But Z OCA + Z OCB = 2 rt. A Hyp. 

.\ZOCA + ZOCF=ZOCA + ZOCB. Ax. 1 

Take away from each of these equals the common Z OCA. 

Then Z OCF= Z OCB. Ax. 3 

.•. CB and (7i^ coincide. 

.*. AC SiXid CB are in the same straight line. at a 

04. Scholium. Since Propositions II. and III. are true, 
their opposites are true ; namely, § 80 

J^ the exterior sides of two adjacent angles are not in a 
straight line, these angles are not supplements of each other. 

If two adjacent angles are not supplements of each other, 
ihsir exterior sides are not in the same straight line. 



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20 



PLANE GEOMETRY. — BOOK 1. 



Proposition IV. Theorem. 

05. tf one straight line intersects another straight 
line, the vertical angles are equal. 




§90 
§90 



Let line OP cut AB at C. 
To prove Z. OCB = AACP. 

Proof. Z OCA + Z0CB = 2rt.A, 

(being sup.-adj. ^). 

Z OCA + Z ACF= 2 rt. A, 
(being sup. -adj. ^). 

.'.ZOCA + ZOCB = ZOCA + ZACF. Ax. 1 

Take away from each of these equals the common Z. OCA. 

Then ZOCB==ZACF. Ax. 3 

In like manner we may prove 

ZACO = ZPOB. aE.D. 

96. Cor. If (me of the four angles formed hy the intersection 
of two straight lines is a right angle, the other three angles are 
right angles. 



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THE STRAIQHT LINE. 21 



Proposition V. Theorem. 

97. From, a point without a straight line one per^ 
pendicuiar, and only one, can he drawn to this line. 

p 
A 



n 



\ 



-B 



P' 
Let P be the point and AB the line. 

To prove that one perpendicular, and only one, can be drawn 
from F to AB. 

Proof. Turn the part of the plane above AB about AB as 
an axis until it falls upon the part below AB, and denote by 
P' the position that F takes. 

Turn the revolved plane about AB to its original position, 
and draw the straight line FF', cutting AB at C. 

Take any other point D in AB, and draw P-D and F'D, 

Since FCF' is a straight line, FDP is not a straight line. 
{Between two points only one straight line can be dravm.) 

.'. Z FCF'is a st. Z, and Z FDF is not a st. Z. 
Turn the figure FCD about AB until F falls upon P. 
Then OF will coincide with OF, and BF with BF. 
.-. Z FOB = Z FOB, and Z FBO= Z FBO. § 59 

.-. Z FOB, the half of st. Z FCF, is a rt. Z ; and Z FBO, 
the half of Z FBF, is not a rt. Z. 

.-. FOis ± to AB, and FB is not J. to AB. § 47 

.•. one X, and only one, can be drawn from Pto AB. 



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22 PLANE GEOMETRY. — BOOK I. 

Parallel Lines. 

98. Dbp. Parallel lines are lines which lie in the same 
plane and do not meet however far they are prolonged in both 
directions. 

99. Parallel lines are said to lie in the same direction when 
they are on the same side of the straight line joining their ori- 
gins, and in opposite directions when they are on opposite sides 
of the straight line joining their origins. 

Proposition VI. 

100. Two straight lines in the same plane perpen- 
dicular to the same straight line are parallel. 



-B 



Let AB and CD he pezpendicnlar to AC. 

To prove AB and CD parallel. 

Proof. If AB and CD are not parallel, they will meet if 

sufficiently prolonged, and we shall have two perpendicular 

lines from their point of meeting to the same straight line ; 

but this is impossible. § 97 

{From a given point without a straight line, one perpendicular, and only 
one, can he drawn to the straight line.) 

.*. AB and CD are parallel. ato. 

Remark. Here the supposition that AB and CD are ru)t parallel leads 
to the conclusion that two perpendiculars can be drawn from a given 
point to a straight line. The conclusion is false, therefore the supposi- 
tion is false; but if it is false that AB and CD are not parallel, it is true 
that they are parallel. This method of proof is called the indirect 
method. 

101. Cor. Through a given point, one straight line, and only 
one, can be drawn parallel to a given straight line. 



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PARALLEL LINES. 



Proposition VII. Theorem. 

102. If a straight line is perpendicular to one of 
two parallel lines, it is perpendicular to the other. 



H 
O 



E 



-IT 



Let AB and EF be two parallel lines, and let HK he 
perpendicular to AB, and cut EF at C 

To prove HKLEF. 

Proof. Suppose MN drawn through C_L to HK, 

Then MN'iQ \\ to AB, § 100 

(two lines in the Bame plant JLto a given line are parallel). 

But ' UF ia\\ to AB. Hyp. 

.-. J^Ji^ coincides with MN, § 101 

{through the same point only one line can be drawn \\ to a given line). 

.'.FFis±to HK, 

that is, B'Kia±to FF. aE.D. 

103. If two straight lines AB 
and CD are cut by a third line 
FF, called a transversal, the 
eight angles formed are named 
as follows : 

The angles a, d, /, g are called 
interior; b, c, e, h are called ex- 
terior angles. 

The angles d and/, or a and g, are called alt. -int. angles. 

The angles h and A, or c and e, are called aU.-ext. angles. 

The angles b and /, c and g, a and e, or d and A, are called 
ext.'int. angles. 




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24 PLANE GEOMETBY. — BOOK I. 



Proposition VIII. Theorem. 

104. If two parallel straight lines are cut hy a third 
straight line, the alternate-interior angles are equal. 




Let EF and GH be two parallel straight lines cut hy 
the line BO. 

To prove ZB^AC. 

Proof. Through 0, the middle point of BC^ suppose AD 
drawn ± to OH, 

Then . AD is likewise JL to EF, § 102 

(a straight line ± to one of two Ms is ± to the other), 

that is, OD and BA are both ± to AD, 

Apply figure COD to figure BOA, so that OD shall fall 
on OA, 

Then OCwill fall on OB, § 95 

(since Z COD =- Z. BOA, being vertical A) ; 

and the point will fall upon B, 

{since 00 = OB by construction). 

Then the ± CD will coincide with the ± BA, § 97 

(Jrom a pclnt toithout a straight line only one ± to that line can be drawn), 

/. Z OCD coincides with Z OB A, and is equal to it. §59 
a. E. D. 

Ex. 1. Find the value of an angle if it is doable its complement ; if 
it is one-fourth of its complement. 

Ex. 2. Find the value of an angle if it is double its supplement ; if it 
is one-third of its supplement. 



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PARALLEL LINES, y 25 



Proposition IX. Theorem. 

105. Conversely : WTien two straight lines are cut 
by a third straight line, if the alternate-interior an- 
gles are equal, the two straight lines are parallel. 




N 



Let EF cut the straight lines AB and CD in the points 
H and K, and let the /.AHK^^ZHKD. 

To prove ' ABWtoCD. 

Proof. Suppose MN drawn through ^ II to CZ) ; § 101 

then Z MHK= A HKD, % 104 

(bdng alt Ant AojW lines). 

But ZAHK=ZHKD. Hyp. 

.-. Z MHK= Z ASK. Ax. 1 

/• the lines JOT and AB coincide. 

But MNi& II to CD. Cons. 

/. AB^ which coincides with MN, is II to OD, 



Ex. 3. How many degrees in the angle formed by the hands of a 
clock at 2 o'clock ? 3 o'clock? 4 o'clock? 6 o'clock? 



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PLANE GEOMETRY. — BOOK I. 



Proposition X. Theorem. 

106. If two parallel lines are^ cutby a third straight 
line, the exterior-interior angles are equal. 




Let AB and CD he two parallel lines cut by the 
straight line EF, in the points H and K» 

To prove Z EHB = Z HKD, 

Proof. AEHB = Z.AHK, §95 

{being vertical A), 

• But AAHK=Z.BKD, §104 

{})dng alt.'int. AqfW lines), 

.\ZEHB = ZHKD. Ax. 1 

In like manner we may prove 

ZEHA = ZHKa 

aE.D. 

107. Cor. The aiternate-exterior angles EHB and CKF, 
and also AHE and DKF, are equal. 



Ex. 4. If an angle is bisected, and if a line is drawn through the 
vertex perpendicular to the bisector, this line forms equal angles with 
the sides of the given angle. 

Ex. 5. If the bisectors of two adjacent angles are perpendicular to 
each other, the adjacent angles are supplementary. 



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PARALLEL LINES. 27 



Proposition XI. Theorem. 

108. Conversely : WTien two straight lines are cwt 
by a third straight line, if the exterior-interior arv- 
gles a/re equal, these two straight lines are parallel. 




Let EF cut the straight lines AB and CD in the points 
H and K, and let the /.EHB = /.HKD. 

Tq prove AB II to CD. 

Proof. Suppose J/iV drawn through S II to OD. § 101 

Then Z USJSr= Z HKD, § 106 

Q>dng ext-int. AofW lines). 

But Z EHB = Z HKD. Hyp. 

.\ZEHB = ZEHN. Ax. 1 

/.the lines JOT and AB coincide. 

But MN'isWioCD, Cons. 

.'. ABt which coincidies with MN, is D to CD. 

aE.D. 



Ex. 6. The bisector of one of two vertical angles bisects the other. 
Ex. 7. The bisectors of the two pairs of vertical angles formed by 
two intersecting lines are perpendicular to each other. 



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28 PLANE GEOMETRY. — BOOK I. 



Pboposition XII. Theorem. 

109, If two parallel lines are cut by a third straight 
line, the sum of the two interior angles on the same 
side of the transversal is equal to two right angles. 




Let AB and CD be two parallel lines cut by the 
straight line EF in the points H and K. 

To prove Z BHK+ Z HKD = 2 rt A 

Proof. ZUB'B + ZBB'K=2Tt.A, §90 

^ {being sup.'adj. A). 

But AEHB=^AHKD, §106 

{})dng ext.-int AofW lines). 

Substitute Z HKD for Z UB'B in the first equality ; 
then Z BIIK+ Z HKD = 2 rt. A, 

Q.E.D. 



Ex. 8. If the angle AHE is an angle of 135®, find the number of 
degrees in each of the other angles formed at the points JEf and K. 

Ex. 9. Find the angle between the bisectors of adjacent complemea- 
tary angles. 



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PARALLEL LINES. 29 



Pboposition XIII. Theorem. 

110. Conversely : WTien two straight lines are cut 
by a third straight line, if the two interior angles on 
the same side of the transversal are together equal to 
two right angles, then the two straight lines are 
paralleL. 




F 

Let EF cut the strsbight lines AB and CD in the points 
H and K, and let the ZBHK + ZMKIJ equal two right 
angles. 

To prove AB II io CD. 

Proof. Suppose -3fiV drawn through ^11 to CD. 

Then Z JSrJ3:K+ Z HKD = 2 rt. A, § 109 

{being two interior A of We on the same side of the transversal). 

But Z.BHK+AHKL = 2Tt,A, Hyp. 

.\ZNHK+AHKD = ZBIIK+AHKD. Ax. 1 

Take away from each of these equals the common Z HKD] 

then Z NHK= Z BHK. Ax. 3 

/. the lines AB and JfiV coincide. 

But JOT is II to CD. Cons. 

.*. AB, which coincides with MN, is II to CD. 

Q.E.a 



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30 



PLANE GEOMETRY. — BOOK I. 



Proposition XIV. Theorem. 

111. Two straight lines which are parallel to a third 
straight line are parallel to ea^h other. 



H 

A j2 B 

C 1£ £> 

E • J- 

K 



Let AB and CD be pajrallel to EF. 

To prove AB W to CD. 

Proof. Suppose JIK drawn ± to EF. § 97 

Since CD and ^i^are II, ^^is ±to CD, § 102 
(if a straight line is ±to one of two lU, it is ± to the other also). 

Since AB and ^i^are II, ^^is also ± to AB. § 102 

.\ZROB = Z]IFD, 

(each being a rt, Z), 

.'. AB is II to CD, § 108 

(when two straight lines a/re cut by a third straight line, if the extAnt. A 
are equal, the two lines are paraUel), 

Q.E.IX 

Ex. 10. It has been shown that if two parallels are cut by a trans* 
versal, the alternate-interior angles are equal, the exterior-interior angles 
are equal, the two interior angles on the same side of the transversal are 
supplementary. State the opposite theorems. State the converse theo- 
rems. 



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PARALLEL LINES. 31 



Peoposition* XV. Theorem. 

112. Two angles whose sides are parallel, ea^ih to 
each, are either equal or supplementary. 




-N 



U 

F 

Let AB be parallel to EF, and BC to MN. 

To prove Z ABO equal to Z EHN, and to Z MHF, and 
mpplementary to Z EHM and to Z NHF, 

Proof. Prolong (if necessary) J? (7 and -F^ until they inter- 
sect at D, § 81 (2) 

Then ZB = ZEBQ, §106 

and Z BHN= Z EDO. § 106 

{pdng ext.-int. AofW lines)^ 

.\ZB = ZBHN\ Ax. 1 

and ZB = Z MHF (the vert. Z of BHN), 

Now Z DHN\a the supplement of Z ERMsind Z NHF. 

.'. Z B, which is equal to Z BHN, 

is the supplement of Z EITMsmd of Z NHF. 

Q.E.D. 

Remabk. The angles are equal when both pairs of parallel sides 
extend in the same direction, or in opposite directions, from their ver- 
tices ; the angles are mpplementary when two of the parallel sides extend 
in the same direction, and the other two in opposite directions, j&om their 
verticea. 



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82 PLANE GEOMETRY. — BOOK I. 



Proposition XVI. Theorem. 

113. Two angles whose sides are perpendicular, each 
to ea^h, are either equal or supplementary. 

G 
K \ 

\ 
\ 



Let AB be perpendicular to FD, and AC to GI. 

To prove Z BAC eqical to Z DFO, and supplementary to 
ZDFL 

Proof. Suppose ^^ drawn ± to AB, and AH 1. to AC, 

Then ^JT is II to i^i>, and ^JT to 7G^, §100 

{two lines -L to the same line are paraUel), 

,\ADFG = AKAH, §112 

ipwo angles are equal whose sides are II and extend in the same direction 
from their vertices). 

The Z BAKi^ a right angle by construction. 

.*. Z BAH'iQ the complement of Z KAH. 
The Z CAII\% a right angle by construction. 

.*. Z BAH\% the complement of Z BAC, 

,\ABAG^AKAH, §87 

(complements of equal angles are equal). 

.\ADFG^ABAC. Ax. 1 

•% Z BFI, the supplement of Z i)i^(?, is also the supplement 

Qi/-BAC Q.E.D. 

Remark. The angles are equal if both are acute or both obtuse ; they 
are supplevientary if one is acute and the other obtuse. 



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PERPENDICULAR AKD OBLIQUE LINES. 



33 



Perpendicular and Oblique Lines. 

Proposition XVII. Theorem. 

114. The perpendicular is the shortest Line that can 
be drawn from a point to a straight Litis, 




Let AB he the given straight line, P the given point, 
PC the perpendicular, and PD any other line drawn 
from P to AB. 

To prove PC < PD. 

Proof. Produce PC to P\ making CP^= PC; and draw I)P\ 
On AB as an axis, fold over CPD until it comes into the 
plane of CP'B. 

The line CP will take the direction of CP\ 

(nnce Z PCD = Z PCD, each being art A by hyp). 

The point P will fall upon the point P', 

{since PC= PC by com.). 

.-.line PZ) = line P'i>, 

.'.PI) + P'D = 2PI), 

and PC + CP' = 2 PC Cons. 

But PC + CP' < PD^DP\ 

(a straight line is the shortest distance between two points). 

.'.2PC< 2PD, or P0< PD. ata 



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S4 PLANS GBOMETBY. — BOOS X. 

llSi Scholium. The distance of a point from a line is under- 
stood to mean the length of the perpendicular from the point 
to the line. 

Proposition XVIII. Theorem. 

116. Two obliqiue lines drawn from a point in a 
-perpendicular to a given line, cutting off equal dis- 
tances from the foot of the perpendicular, are equal. 




Let FO be the perpendicular, and CA and CO two 
oblique lines cutting off equal distances from F. 

To prove CA==CO. 

Proof. Fold over CFA, on CFas an axis, until it comes into 
the plane of CFO, 

FA will take the direction of FO, 
{since A CFA = Z CFO, each being art. A by hyp.). 

Point A will fall upon point 0, 
{since FA = FO by hyp.). 

.-. line CA = line CO, 
{their extremities being the same points), a B. a 

117. Cor. Two oblique lines dravm from a point in a per- 
pendicular to a given line, cutting off eqvxil distances from the 
foot of the perpendicular, make eqiuil angles with the given line^ 
and also with the perpendicular. 



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PEBPENPIOULAE AND OBLIQUE LINES. S5 



Proposition XIX. Theorem. 

118i The sum of two lines drawn from a point to 
the extremities of a straight line is greater than the 
sum of two other lines similarly drawn, hut included 
hy HufWu 

C 




Let CA and CB be two lines drawn from the point C 
to the extremities of the straight line AB. Let OA and 
OB be two lines similarly drawn, but included by CA 
and CB. 

To prove CA+CB>OA + OB. 

Proof. Produce AO to meet the line CB at U. 

Then AC+ Ci:>OA + OF, 

(a itraight line is the shortest distance between two poinU), 

and BE+OE> BO. 

Add these inequalities, and we have 

CA+CE+BE+OE> OA + OE+OB. 

Substitute for CE+ BE its equal CB, 

and take away OE from each side of the inequality. 

We have CA + CB>OA + OB. Ax. 5 ^^.b^ 



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86 



PLANE GEOMETRY. — BOOK I. 



Proposition XX. Theorem. 

119. Of two oblique lines drawn from the same 
-point in a perpendicular, cutting off unequal dis- 
tances from the foot of the perpendicular, the more 
remote is the greater. 






Let OC be perpendicular to AB, OG and OE two oblique 
lines to AB, and CE greater than CG. 

To prove OE>OG, 

Proof. Take CF equal to CO, and draw OF, 

Then OF=OQ, §116 

(^1^0 oblique lines drawn from a point in a ±, cutting off equal didances 
from the foot of the ±, are equal). 

Prolong OC io D, making CD =00, 

Draw ED and FD, 

Since AB is -L to OD at its middle point, 

FO = FD, and FO = FD, 

But OF+FD> 0F+ FD, 

{the sum of two oblique lines drawn from a point to the extremities of a 
straight line is greater than the sum of two other lines similarly drawn, 
but included by them). 

.'. 20F>20F, or 0F> OF 
But 0F= 00. Hence OF > 00. q. e. d. 

120. Cor. Only (wo equal straight lines can be drawn from 
a point to a straight line ; and of two unequal lines, the greater 
cuts off the greater distance from the foot of the perpendicular. 



§116 
§118 



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PERPENDICULAR AND OBLIQUE LINES. 



37 



Proposition XXI. Theorem. 

121* Two equal obliovs lines, drawn from the same 
point in a perpendicular, cut off equal distances from 
the foot of the perpendicular. 




Let CF be the perpendicular, and CE and OK be two 
equal oblique lines drawn from the point C to AB. 

To prove FE=FK 

Proof. Fold over CFA on CFas an axis, until it comes into 
the plane of CF£. 

The line FE will take the direction FK, 
(since Z. CFE^ Z CFK, each being a rt. Z. hy hyp.). 

Then the point F must fall upon the point K, 

and FF=FK. 

Otherwise one of these oblique lines must be more remote 
from the perpendicular, and therefore greater than the other ; 
which is contrary to the hypothesis that they are equal. § 119 

Q.E.D. 



Ex. 11. Show that the bisectors of two Bupplementary-adjacent 
angles are perpendicular to each other. 

Ex. 12. Show that the bisectors of two vertical angles form one 
straight line. 

Ex. 13. Find the complement of an angle containing 26° 62^ 37^^ 
Find the supplement of the same angle. 



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38 



PLANE GEOMETRY. — BOOK L 



PiROPOsiTioN XXII. Theorem. 

122. Every point in the perpendicular, erected cut 
the middle of a given straight line, is equidistant 
from the extremities of the line, and every point not 
in the perpendicular is unequally distant from the 
extremities of the line. 




Let PR be b perpendicular erected at the middle of 

the straight line AB, any point in PR, and C any 

point without PR. 

Draw OA and OB, CA and CB. 

To prove OA and OB equal, CA and CB unequal. 

Proof. FA = FB. Hyp. 

.'.OA^OB, §116 

{two oblique lines drawn from the same point in a JL, cutting off equal dis- 
tances from the foot of the ±, are equal). 

Since C is without the perpendicular, one of the lines, OA 

or CB, will cut the perpendicular. 

Let CA cut the ± at D, and draw DB. 

Then DB = DA, §116 

if/iDO oblique lines drawn from the same point in a ±, cutting off equal dii- 
tancesfrom the foot of the ±, are equal). 

But C£<CI>+DB, 

{a straight line is the shortest distance between two points). 

Substitute in this inequality DA for BB, and we have 

CB<CI> + I)A. 
That is, CB < CA. ^^o. 



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PERPENDICULAR AND OBLIQUE LINES. 39 

123. Since two points determine the position of a straight 
line, two points equidistant from the extremities of a line deter- 
mine the perpendicular at the middle of that line. 

The Locus of a Point. 

124. If it is required to find a point which shall fulfil a 
single geometric condition, the point will have an unlimited 
number of positions, but will be confined to a particular line, 
or group of lines. 

Thus, if it is required to find a point equidistant from the 
extremities of a given straight line, it is obvious from the last 
proposition that an7/ point in the perpendicular to the given 
line at its middle point does fulfil the condition, and that no 
other point does ; that is, the required point is confined to this 
perpendicular. Again, if it is required to find a point at a 
given distance from a fixed straight line of indefinite length, it 
is evident that the point must lie in one of two straight lines, 
so drawn as to be everywhere at the given distance from the 
fixed line, one on one side of the fixed line, and the other on 
the other side. 

The hcu^ of a point under a given condition is the line, 
or group of lines, which contains all the points that fulfil the 
given condition, and no other points. 

125. Scholium. In order to prove completely that a certain 
line is the locus of a point under a given condition, it is neces- 
sary to prove that every point in the line satisfies the given 
condition; and secondly, that every point which satisfies the 
given condition lies in the line (the converse proposition), or 
that every point not in the line does not satisfy the given condi- 
tion (the opposite proposition). 

126. CoR. The locus of a point equidistant from the extrem- 
ities of a straight line is the perpendicular bisector of that line. 

§§ 122, 123 



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40 



PLANE GEOMETBY. — BOOK I. 



Triangles. 

127. A triangle is a portion of a plane bounded by three 
straight lines ; as, ABC. 

The bounding lines are called the 
sides of the triangle, and their sum is 
called it& petimeter ; the angles formed 
by the sides are called the angles of the 
triangle, and the vertices of these an- 
gles, the vertices of the triangle. 

128. An exterioi* angle of a triangle 
is an angle formed between a side and 
the prolongation of another side; as, 
ACD. The interior angle ACB is 
adjacent to the exterior angle ; the "" Fig. 2. 
other two interior angles, A and B, are called opposite^ 
interior angles. 




B 




Scalene. 



Isosceles. 



Equilateral. 



129. A triangle is called, with reference to its sides, a 
scalene triangle when no two of its sides are equal ; an isos- 
celes triangle, when two of its sides are equal ; an equilateral 
triangle^ when its three sides are equal. 





Right Obtuse. Acute. Equiangular. 

130. A triangle is called, with reference to its angles, a right 
triangle^ when one of its angles is a right angle ; an obtuse 



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TEIANGLES. 41 

triangle^ when one of its angles is an obtuse angle ; an acute 
triangle, when all three of its angles are acute angles ; an 
equiangular triangle, when its three angles are equal. 

131. In a right triangle, the side opposite the right angle is 
called the hypotenuse, and the other two sides the legs, of the 
triangle. 

132. The side on which a triangle is supposed to stand is 
called the ha^se of the triangle. Any one of the sides may be 
taken as the base. In the isosceles triangle, the equal sides 
are generally called the legs, and the other side, the base, 

133. The angle opposite the base of a triangle is called the 
vertical angle, and its vertex the vertex of the triangle. 

134. The altitude of a triangle is the perpendicular distance 
from the vertex to the base, or to the base produced ; as, AI>, 

135. The three perpendiculars from the vertices of a tri- 
angle to the opposite sides (produced if necessary) are called 
the altitudes; the three bisectors of the angles are called tha 
bisectors; and the three lines from the vertices to the middle 
points of the opposite sides are called the medians of the 
triangle. 

136. If two triangles have the angles of the one equal respec- 
tively to the angles of the other, the equal angles are called 
homologous angles, and the sides opposite the equal angles are 
called homologous sides. 

In general, points, lines, and angles, similarly situated in 
equal or similar figures, are called homologous, 

137. Theorem. The sum of two sides of a triangle is greater 
than the third side, and their difference is less than the third 
side. 

In the A ABO {Yig, 1), AB+BC>AO, for a straight line 
is the shortest distance between two points ; and by taking 
away 5(7 from both sides, AB>AO-BC, or AO-BC<AB, 



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42 PLANE GEOMETRY. — BOOK I. 



Proposition XXIII. Theorem. 

138. The sum of the three angles of a triangle is 
equal to two right angles. 



A c """F 

Let ABC be a triangle. 

To prove Z£ + Z BCA + ZA = 2 rt. A 

Proof. Suppose CE drawn II to AB, and prolong AC to F. 

Then Z ECF+Z UCB + Z BCA = 2rt A, § 92 

{the turn of aU the A about a point on the same aide of a straight line 

But ZA = ZECF, §106 

Q>eing extAnt, AofW lines). 

sixid ZB^ZjBCF, §104 

(being alt-int AofW lines). 

Substitute for Z FCFmd Z BCE the equal A A and B. 

Then Z A+Z B + Z BCA = 2Ti. A. 

139. Cor. 1. If the sum of two angles of a triangle is sub- 
tracted front two right angles, the remainder is eqital to the 
third angle. 

140. Cor. 2. If two triangles have two angles of the one 
eqtuil to two angles of the other, the third angles are equal. 

141. Cor. 3. if two right triangles have an acute angle of 
the one equal to an aciUe angle of the oiher^ the other acute 
angles are equal. 



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TRIANGLES. 43 

142. Cor. 4. in a triangle there can be but one right angle^ 
or one obticse angle. 

143. Cor. 6. In a right triangle the two acute angles are 
complements of each other. 

144. Cor. 6. In an equiangular triangle, each angle is one- 
third of two right angles^ or two-thirds of one right angle. 



Proposition XXIV. Theorem. 

'145. The exterior angle of a triangle is equal to the 
sum of the two opposite interior angles. 




Let BCH be an exterior angle of the triangle ABC. 

To prove Z. BQH= ZA + ZB. 

Proof. Z £C£r+ ZACB = 2vt A, 

{being sup. -adj. A). 

ZA + ZB + ZACB = 2rtA, §138 

(the sum of the three AofaA = 2rt. A). 

.\ZBOH+ZACB = ZA + ZB + ZACB. Ax. 1 

Take away from each of these equals the common Z ACB ; 

then ZBCH^ZA + ZB. Ax. 3 

aE.D. 

146. Cor. The exterior angle of a triangle is greater than 
either of the opposite interior angles. 



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44 PLANE GEOMETRY. — BOOK I. 



^ Proposition XXV. Theorem. 

147. Two triangles are equal if a side and two ad- 
jacent angles of the one are equal respectively to a 
side and two adjax^nt angles of the other. 





A c JD F 

In the triangles ABC and DBF, let AB = DE, ^A-^D, 

To prove A ABC= A DEF, 

Proof. Apply the A ABC to the A DEF so that AB shall 

coincide with DE. 

« 

-4 C will take the direction of DF, 
{for/.A^/.D,hyhyp)\ 

the extremity Cof ^C will fall upon DF or DF produced. 

BO will take the direction of EF, 
(for ZB''ZE,by hyp.) ; 

the extremity (7 of ^(7 will fall upon EF or -EF produced. 

.*. the point (7, falling upon both the lines DF and EF^ 
must fall upon the point common to the two lines, namely, F. 

.'•the two A coincide, and are equal. aE.D. 

148. Cor. 1. Two right triangles are equal if the h^potehuse 
and an acute angle of the one are equal respectively to the hypote^ 
nuse and an acute angle of the other, 

149. Cor. 2. Two right triangles are equxil if a side and an 
acute angle of the one are equal respectively to a side and 
homologous acute angle of the other* 



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TRIANGLES. 46 



Pboposition XXVI. Theoeem. 

160. Two triangles are equal if two sides and the 
included angle of the one are equal respectively to 
two sides and the included angle of the other. 





In the triajigles ABC and DBF, let AB = DE, AC= DF, 
ZA = /:d, 

To prove AA£C-=A DEF. 

Proof. Apply the A ABC to the A DEF so that AB shall 
coincide with BE. 

Then -4 (7 will take the direction of DF, 

{forZA=^ZD,byhyp.); 

the point C will fall upon the point F, 
(JorAC=DF, by hyp.). 

.\CB = FE, 
{their extremities being the same points). 

•*. the two A coincide, and are equal. 

151. Cor. Two right triangles are equal if their legs are 
eqv^lj each to each. 



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46 PLANE GEOMETRY. — B(X)K I. 



Proposition XXVII. Theorem. 

152. If two triangles have two sides of the one equal 
respectively to two sides of the other, but the included 
angle of the first greater than the included angle of 
the second, then the third side of the first will be 
greater than the third side of the second. 




In the triangles ABC and ABE, let AB-AB, BC=BE; 
but A ABO greater than /.ABE. 

To prove A0> AE. 

Proof. Place the A so that AB of the one shall coincide with 
AB of the other. 

Suppose J?i^ drawn so as to bisect Z, EBG. 
Draw EF. 

In the A JE^^i^'and CBF 

EB = BC, Hyp. 

BF=BF, Iden. 

Z,EBF=ACBF. Cons. 

.-. the A EBF and QBE are equal, § 150 

{haxing two aides and the included A of one equal respectively to two sides 
and the included Z of the other). 

.'. EF= EC, • 
(being homologous sides of equal A). 

Now AF+FE>AIJ, §187 

{the turn of two gides ofaAii greater than the third tide). 
.\AF+FO>AE; 

or, AO>AE. Q.E.O. 



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TRIANGLES. 47 



Peoposition XXVIII. Theorem. 

153. Conversely. If two sides of a triangle are equal 
respectively to two sides of another, hut the third side 
of the first triangle is greater than the third side of 
the second, then the angle opposite the third side of 
the first triangle is greater than the angle opposite 
the third side of the second. 

D 





B c E 

In the triangles ABC and DEF, let AB=DE, AC = DF; 
but let BC be greater than EF. 
To prove A A greater than A D. 

Proof. Now Z J. is equal to Z. Z>, or less than Z Z), or 
greater than Z D. 

But Z ^ is not equal to Z Z>, for then A ABC would be 
equal to A DEF, § 150 

{JumTtg two sides and the included Z of the one respectively equal to two 
sides and the included Z of the other), 

and ^C would be equal to UF. 

And Z ^ is not less than Z I), for then BO would be less 

than FF § 162 

.*. Z ^ is greater than Z D, 

as-a 



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48 PLANE GEOMETEY. — BOOK L 



Peoposition XXIX. Theoeem. 

154. In an isosceles triangle the angles opposite the 
equal sides are equal, 

A 



B D C 

Let ABO be an isosceles triangle, having the sides 
AB and AC equal. 

To prove Z.B = /.C. 

Proof. Suppose AD drawn so as to bisect the ABAC, 

In the A ADB and ADO^ 

AB = Aa Hyp. 

AD = AD, Men. 

j^BAD = Z,CAD. Cons. 

.\ A ADB ^ A ADO, §150 

{two 4 are equal if two sid^s and the included Z of the one are equal 
respectively to two sides and the included Z of the other\ 

•* • Z. B "=- zL Q, 0. E. D. 

155. Cor. An equilateral tnangle is equiangular, and each 
angle contains 60°. 

Ex. 14. The bisector of the vertical angle of an isosceles triangle 
bisects the base, and is perpendicular to the base. 

Ex. 15. The perpendicular bisector of the base of an isosceles triangle 
passes through t^e vertex and bisects the angle at the vertex. 



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O^IANGLES. 



•49 



Proposition XXX. Theorem. 

166. If two angles of a triangle are equal, the sides 
opposite the equaZ angles are equal, and the triangle 
is isosceles. 

A 




In the triangle ABC, let the Z.B=^0, 
To prove AB=^AC. 

Proof. Suppose AD drawn ± to BC. 

In the rt. A ADB and ADC, 

AD = AD, Iden. 

ZB = Za Hyp. 

.-. rt. A ADB = rt. A ADC, § 149 

(having a side and an acute Z of the one equal respectively to a side and 
an homologous acute Z. of the other). 



.\AB = AC, 
(being homologous sides of equal ^). 

167. Cor. An equiangular triangle is also equilatei'al. 



Q.E.O. 



Ex. 16. The perpendicular from the vertex to the base of an isoscelee 
triangle is an axis of symmetry. 



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60 PLAKE GEOMETfiy. — BOOK 1. 



Fboposition XXXI. Theobeh. 

IBB, If two sides of a triangle are unequal, the an- 
gles opposite are unequal, and the greater angle is 
opposite the greater side. 




C B 

In the triangle ACB let AB be greater tli&n AC. 

To prove A A CB greater than Z B. 

Proof. Take AE equal to AO. 

Draw EC, 

ZAEC=ZACE, §154 

{being A opposite equxd sides). 

But ZAEC'iQ greater than Z B, § 146 

(an exterior A of a t^ is greater than either opposite interior /). 

and Z ACB is greater than Z ACE. Ax. 8 

Substitute for ZACE\\;& equal Z AEO, 

then Z ACB is greater than Z AEC 

Much more, then, is the Z ACB greater than Z B» 

aE.a 



Ex. 17. If the angles il^Cand ACB, at the base of an isosceles tri- 
angle, be bisected by the straight lines BD, CD, show that D^Cwill 
be an isosceles triangle. 



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TRIANGLES. 61 



Peoposition XXXII. Theobem. 

' 159. Conversely : If two angles of a triangle are 
unequal, the sides opposite are unequal, and the 
grealer side is opposite the grealer angle. 




In the triangle ACB, let angle ACB be greater tZuui 
angle B. 
To prove AB > AG. 

Proof. Now AB is equal to AG, or less than AG, or greater 
than AG. 

But AB is not equal to AG, for then the Z G would be 
equal to the Z, B, § 154 

Qidng A opposite equal sides). 

And AB is not less than AG, for then the Z G would be 
less than the Z B, § 158 

(*^ two tides of a A are unequal, the A opposite are unequal, and the 
greater Z is opposite the greater side). 

/. AB is greater than AG. 

Q.E.D. 



Ex. 18. ABO And ABB are two triangles on the same base AB, and 
on the same side of it, the vertex of each triangle being Mfithout the 
other. If ^C equal AD, show that BC cannot equal 
BD. 

Ex. 19. The sum of the lines which join a point 
within a triangle to the three vertices is less than 
the perimeter, but greater than half the perimeter. ^ 




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52 



PLANE GEOMETRY. — BOOK I. 



Peoposition XXXIII. Theorem. 

160. Two triangles are equal if the three sides of 
the one are equal respectively to the three sides of 
the other. 





In the triangles ABC and A^Bfa, let AB = A'B^, AC^A'C, 
BC=B'a. 

To prove A ABC= A A*BQK 

Proot Place A A^B^O^ in the position AffO, having its 
greatest side -4'C in coincidence with its equal AC, and its 
vertex at £f, opposite B ; and draw BB*. 

Q\nQQAB = AB\ Hyp. 

AABB'^AAB'B, §154 

(in an isoKdes A the A opposite the equal tides are equal), 

Since CB= Off, Hyp. 

ZOBB' = ZCB^B. §154 

Hence, Z ABO= Z ABfG, Ax. 2 

/. A ABC= A AB'0=- A A^B/C^ § 150 

{f,yoo ^ are eqml if two tides and included Z of one are equal to two 
sides and included /.of the oHket), 



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TRIANGLES. 



63 



Proposition XXXIV. Theorem. 

161. Two right triangles are equal if a side and 
the hypotenuse of the one are equal respectively to a 
side and the hypotenuse of the other. 




B'^ 




It the right triangles ABC and A'S^a, let AB^A^B, 
and AC=A'C. 



To prove 



AA£C=^AA'B'C'. 



Proof. Apply the A ABC to the A A^B^C, so that AB shall 
coincide with A'B', A falling upon A\ B upon B', and (7 and 
(T upon the same side of A'B\ 

Then BO will take the direction of ffO^, 

{for Z -45C- Z A'SfQ^, each being aH.^ 
Since AC^A'0\ 

the point C will fall upon C, § 121 

(two e^jwd obUque lines from a point ina ±ciU off equal distances from 
the foot of the ±), 



•*• the two A coincide, and are equal. 



Q.i.a 



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64 PLANE GEOMETRY. — BOOK I. 



Proposition XXXV. Theorem. 

162. Every point in the bisector of an angle is equi- 
distant from the sides of the angle* 






Let AD be the bisector of the angle BAG, and let O 
be any point in AD. 

To prove that is equidistant from AB and AC, 

Proof. Draw 0-Pand 00 X to AB and -4 (7 respectively. 

In the rt. A J^Oi^ and AOQ 

AO = AO, Iden. 

ZBAO = ZCAO. Hyp. 

,\AAOF=AAOG, §148 

(fwo rL 4 are equal if the hypotenuse and an acute Z of the one are equal 
retipedxvely to the hypotenuse and an acute Z of the other), 

... 0F= 00, 

{homologous sides of equal ^). 
.*. is equidistant from AB and AC. 

Q.I.D. 



What is the locus of a point : 

Ex. 20. At a given distance from a fixed point ? J 57. 

Ex. 21. Equidistant from two fixed points? % 119. 

Ex. 22. At a given distance from a fixed straight line of indefinite 
length ? 

Ex. 23. Equidistant from two given parallel lines ? 

Ex. 24. £(|uidistant from the extremities of a given line? 



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TRIANGLES. 56 



Proposition XXXVI. Theorem. 

163. Every point ivithin an angle, and equidistant 
from its sides, is in the bisector of the angle. 




Let be equidistant from the sides of the angle 
BAG, and let AO join the vertex A and the point 0. 

To prove that AO is the bisector of A BAO. 

Proof. Suppose OF and 00 drawn ± to AB and AC^ 
respectively. 

In the rt. A ^Oi^ and AOG 

0F= 00, Hyp. 

AO=AO. Iden. 

r.AAOF^AAOG, §161 

(two ft. A art equalij the hypotenuse and a side of the one are equal to the 
hypotenuse and a side of the other). 

.'.ZFAO = ZOAO, 
(homologous A of equal ^). 

/. -40 is the bisector of Z BAO, 

aE.a 

164. Cor. The locus of a point within an angle, and equi- 
distant from its sides, is the bisector of the angle. 



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56 PLANE GEOMETEY. — BOOK I- 



QUADBILATEEALS. 

165. A qicadrilcUeral is a portion of a plane bounded by 
four straight lines. 

The bounding lines are the sideSy the angles formed by these 
sides are the angles^ and the vertices of these angles are the 
vertices, of the quadrilateral. 

166. A trapezium is a quadrilateral which has no two sides 
parallel. 

167. A trapezoid is a quadrilateral which has two sides, and 
only two sides, parallel. 

168. A parallelogram is a quadrilateral which has its oppo- 
site sides parallel. 




Trapezium. Trapezoid. Parallelogram. 

169. A rectangle is a parallelogram which has its angles 
right angles. 

170. A rhomboid is a parallelogram which has its angles 

oblique angles. 

171. A square is a rectangle which has its sides equal. 

172. A rhomhus is a rhomboid which has its sides equal. 




Square. Rectangle. Rhombus. Rhomboid. 

173. The side upon which a parallelogram stands, and the 
opposite side, are called its lower and upper hoses. 



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QUADRILATERALS. 57 

174. The parallel sides of a trapezoid are called its bases, 
the other two sides its legSy and the line joining the middle 
points of the legs is called the median. 

176. A trapezoid is called an isosceles trapezoid when its 
legs are equal. 

176. The altitude of a parallelogram or trapezoid is the 
perpendicular distance between its bases. 

177. The diagonal of a quadrilateral is a 
straight line joining two opposite vertices. 




Proposition XXXVII. Theorem. 

178. The diagonal of a paraZleLogram divides the 
figure into two equal triangles. 




Lei ABCE be a parallelogram and AC its diagonal. 

To prove A AJBC= A AUG. 

Inthe A ABO s,nd AUG, 

AC=AC, Iden. 

ZAC£ = ZOAI!, §104 

and ZOAB = ZAOE, 

{being alt.-int, AofW lines), 

.'.AA£C=AAIJfy, §147 

{Tumng a side and two adj. A of the one equal respectively to a sii 
two adj. A of the other). 



side and 

Q.E.O. 



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68 PLANE GEOMETRY. — BOOK I. 



Proposition XXXVIII. Theorem. 



179. In a -paralleio^ram the opposite sides are equal, 
and the opposite angles are equal. 




Let the figure ABCE be a parallelogram. 

To prove B0= AE, and AB = EC, 

oho, ZB = ZE,&ndZ BAE=ZBOE. 

Proof, Draw AC, 

AABC=AAEC, §178 

{ihe diagonal of ad divides the figure into two equal A), 

.'. BC= AE, and AB= CE, 
(being homologous sides of equal A). 

Also, Z ^ = Z ^, and Z BAE= Z BCE, § 112 

(having their sides II and extending in opposite direcUons from 
their vertices), 

Q.E.D. 

180. Cor. 1. Faralkl lines comprehended between parallel 
lines are equal. A B 

181. Cor. 2. Two parallel lines 

are everywhere eqiLally distant, 

For if AB and DC are parallel, 
-h dropped from any points in -4^ to DC, measure the distances 
of these points from DC. But these -h are equal, by § 180 ; 
hence, all points in AB are equidistant from DC 



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QUADRILATERALS. 59 



Proposition XXXIX. Theorem. 

182. If two sides of a qu(idrila;teral are eqimL cund 
parallel, then the other two sides are equal and par* 
allel, and the figure is a parallelogram. 




Let the figure ABCE be a qnadrilateraJ, having the 
side AE equal and parallel to BO. 

To prove AB equal and II to EC. 

Proof. Draw AC. 

In the A ^^(7 and ^^(7 

BC=AE, Hyp. 

AC^AC, . Iden. 

ZBCA^jiCAE, §104 

(being aU,-int. AqfW lines), 

,\AABC=AACE, §150 

{hamng two sides and the included A of the one equal respecUvdy to two 
sides and the included Z of the other). \ 

.•.AB = EO, 

(being homologous sides of equal A). 

Also, ZBAC=ZACE, 

(being homologous A of equal A). 

.-. ^5 is II to ^C, §105 

(when two etraight lines are cut by a third straiaht line, if (he aUAnt A 
are equals the lines are parculet). 

.\ the figure ABCE is a O, § 168 

(the opposite sides being pardUel), q. e. at 



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60 PLANE GEOMETRY. — BOOK I. 



Pbopositioh XL, Theorem. 

183. If the opposite sides of a quadrilateral are 
equal, the figure is a parallelograTrv. 




Let the Hgure ABCE be & quadrilateral having BG^s 
AE and AB = EC. 

To prove figure ABCE a O. 

Proof, Draw -4(7. 

* Inthe A.4-BCan(i^JS7(7 

£0= AE, Hyp. 

A£=OE, Hyp. 

A0= AC. Iden. 

.\AABO=AAEO, §160 

(haioing three tides of the one equal respectively to three sides of the other), 

.'.ZACB==ZCAE, 

and ZBAO^ZACE, 

(being homologous A of equal &^ 

.'.BC'm^tQAE, 

and AB is II to EC, § 105 

{yohen two straight lines lying in the same plane are cut by a third straight 

line, if the alt.-xnt, A are equalt the lines are parallel), 

.-. the figure ABCE is a D, § 168 

{having its opposite sides paralld), 

Q.E.O. 



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QUADRILATERALS. 61 



Propositioji XLI. Theorem. 

184. The diagonals of a parcMelograrrv bisect each 
other. 

B . g 




B 

Let the iignre ABCE be a parallelogram, and let 
the diagonals AC and BE cut each other at 0. 

To prove AO=OC, and BO = OU. 

In the A AOE and BOO 

AE=BO, §179 

{]!)eing opposite sides of a CJ)» 

ZOAI]='ZOCB, §104 

and ZOBA=Z0BO, 

(being cUtAnt. AofW lines), 

,\AAOB = ABOC, §147 

{having a side and two adj. A of the one equal respectively to a side and 
tv9o a^. A of the other). 

.\ A0= 00, a.nd BO =0E, 
(being homologous sides of equal A). 




Ex. 25. If the diagonals of a quadrilateral bisect each other, the figure 
is a parallelogram. 

Ex. 26. The diagonals of a rectangle are equal. 

Ex. 27. If the diagonals of a parallelogram are 
equal, the figure is a rectangle. 

Ex. 28. The diagonals of a rhombus are perpendicular to each other, 
and bisect the angles of the rhombus. 

Ex. 29. The diagonab of a square are perpendicular to each other, 
and bisect the angles of the square. 



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62 PLANE GEOMETRY. — BOOK I. 



Proposition XLII. Theorem. 

186. Two parallelograms, having two sides and the 
included angle of the one eqiiaZ respectively to two 
sides and the included angle of the other, are equal. 



A! 

In the paraUelograms ABCD and A'B'C'D^, let AB:^ 
AfB», AD=A'Df, and ZA = ZA*. 
lb prove thai the [U are equal. 

Apply O ABCD to O A'B^C^iy, so that AD wfll fall on 
and coincide with A*D\ 

Then AB will fall on A^B\ 

(forZA=iZ.A\byhyp.l 

and the point B will fall on B\ 

(forAB^A^B^hyhyp.), 

Now, BO and B'O^ are both B to A'D' and are drawn 
through point B'. 

.-. the lines BO and B'O^ coincide, § 101 

and C falls on B'O' or B'O^ produced. 
In like manner, DO and D'O^ are D to A'B' and are drawn 
through the point 2>^ 

.-. DC and D'C coincide. § 101 

/. the point C falls on D'0\ or D'O^ produced. 

.-. C falls on both B'O' and D*OK 

/. Cmust fall on the point common to both, namely, 0\ 

.'. the two HJ coincide, and are equal. 

Q. E. O. 

186. Cor. Two rectangles having eqiuil bases and equal 
altitudes are equal. 



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QUADRILATERALS. 63 



Proposition XLIII. Theorem. 

187. If three or more parallels intercept equal parts 
an any transversal, they intercept eqiial parts an 
every transversal. 




Let the pare,llels AH, BK, CM, DP intercept equal 
parts HK, KM, MP on the transversal HP. 

To prove that they intercept equal parts AB, BO, CD on the 
transv&rsal AD, 

Proof. From A, B, and (7 suppose AE, BF, and CO drawn 
U to J?P. 

Then AE= HK, BF== KM, CQ = MP, § 180 
{^paralUU comprehended between parallels are equal). 

.:AE=BF=CQ. Ax. 1 

Also Z.A = ZB = AO, §106 

ipeing extAnt. AofW lines) ; 

and AE=Z.F=AQ, §112 

{Jyamig their sides II and directed the same way from the vertices), 

.-. A ABE=A BCF= A CDG, § 147 

(each having a side and two ac^. A respectively equal to a tide and two 
ac^. A of the others), 

.\AB==BC==CD, 
(homologous sides of equal A). % t. a 



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64 PLANE GEOMETRY. — BOOK I. 

188. Cor. 1. The line parallel to the hose of a triangle and 
bisecting one dde bisects the other side also. 
For, let DU be 11 to BC and bisect AB, 
Draw through A a line 11 to J3C. Then 
this line is II to DU, by § 111. The three 
parallels by hypothesis intercept equal 
parts on the transversal AB, and there- ^ 
fore, by §187, they intercept equal parts on the transversal 
AC; that is, the line Z)^ bisects AC, 

189. CoE. 2. The line which joins the ^middle points of two 
sides of a triangle is parallel to the third side, and is equal to 
half the third side. For, a line drawn through 2), the middle 
point oi AB^ H to BC^ passes through -£7, the middle point of 
•^^1 by § 188. Therefore, the line joining I) and U coincides 
with this parallel and is 11 to BC. Also, since UI^ drawn I 
to AB bisects AC, it bisects BC, by § 188 ; that is, BF== FC 
=\ BC. But BDEF is a O by construction, and therefore 
DE=BF=^\Ba 

190. Cor. 3. The line which is parallel to the basses of a trap^ 
ezoid and bisects one leg of the trap- 
ezoid bisects the other leg also. For 
if parallels intercept equal parts on 
any transversal, they intercept equal 
parts on every transversal by § 187. 

191. Cor. 4. The median of a 
trapezoid is parallel to the bases, and is equal to half the sum 
of the bases. For, draw the diagonal DB. In the A ADB 
join F, the middle point of AD, to F, the middle point of DB, 
Then, by § 189, ^i^is II to AB Sini=- ^AB. In the ADBC 
join i^to G, the middle point of BC. Then FG is II to DC 
and = ^DC. AB and FG, being II to DC, are II to each other. 
But only one line can be drawn through F II to AB. There- 
fore FG is the prolongation of FF, Hence FFG is II to AB 
and DC, and = } {AB + DC). 




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EXEBOISES. 66 



Exercises. 

30. The bisectors of the angles of a triangle meet in a point which is 
equidistant from the sides of the triangle. 

Hint. Let the bisectors AD and BE intersect at 0, 
Then being in AD is equidistant from AC and AB. 
(Why ?) And being in BE is equidistant from BO 
and AB. Hence is equidistant from AQ and BG^ 
and therefore is in the bisector CF. (Why ?) 

31. The perpendicular bisectors of the sides of a triangle meet in a 
point which is equidistant from the vertices of the 
triangle. 

HiUT. Let the JL bisectors EE^ and DD^ intersect i>/^ ^i:>^ 

at 0. Then O being in EE^ is equidistant from A 

and C. (Why ?) And O being in DD^ is equidistant r 

from A and B, Hence is equidistant from B and (7, and therefore 

18 in the ± bisector J^JP'. (Why?) 

32. The perpendiculars from the vertices of a* triangle to the opposite 
sides meet in a point. . > a 





Hint. Let the Jl be AH, BP, and CK ^\''"~^'^ 7^ 
Through A, B, suppose B^(y, A^CX, A^B^ \ 
drawn II to BQ, AC, AB, respectively. Then ^\h 

AHi% Xio B'€^. (Why?) Now ABCB^ and \ / 

ACBO' are m (why?), and AB^ = BC, and AO' ^^' 

-» BC. (Why 7) That is, ^1 is the middle point of B^C^. In the same way, 
B and C are the middle points of A^O^ and A^B^, respectively. There- 
fore, AH, BP, and CS'are the ± bisectors of the sides of the A A'B^O^, 
Hence they meet in a point. (Wliy ?) 

33. The medians of a triangle meet in a point which is two- thirds of 
the distance from each vertex to the middle of the opposite side. 

Hint. Let the two medians AD and CE meet in 0. 
Take ^the middle point of OA, and of OC. Join 
QF, FE, ED, and DG. In A^OC, G^i^is II to AC 
and equal to MCI (Why?) i>JS? is II to -4(7 and equal 
to iAC (Why?) Hence DGFE is a O. (Why?) 
Hence AF^ FO = OD, and CQ=00=- OE. (Why ?) ^ 
Hence, any median cuts off on any other median two-thirds of the dis- 
tance from the vertex to the middle of the opposite side. Therefore the 
median from B will cut off AO, two-thirds of AD; that \b, will pass 
through O. 




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66 PLANE GEOMETRY. — BOOK L 



Polygons in General. 

192t K polygon is a plane figure bounded by straight lines. 

The bounding lines are the aides of the polygon, and their 
sum is the perimeter of the polygon. 

The angles which the adjacent sides make with each other 
are the angles of the polygon, and their vertices are the ver- 
ticea of the polygon. 

The number of sides of a polygon is evidently equal to the 
number of its angles. 

193. A diagonal of a polygon is a line joining the vertices 
of two angles not adjacent \ ^ AC, Fig. 1. 





E 

Fio. 1. Fio. 2. FiQ. 3. 

194. An equilateral polygon is a polygon which has all its 
sides equal. 

195. An eqvMmgular polygon is a polygon winch has all its 
angles equal. 

196. A convex polygon is a polygon of which no side, when 
produced, will enter the surface bounded by the perimeter. 

197. Each angle of such a polygon is called a salient angle, 
and is less than a straight angle. 

198. A concave polygon is a polygon of which two or more 
sides, when produced, will enter the surface bounded by the 
perimeter. Fig. 3. 

199. The angle FDE is called a re-entrant angle, and is 
greater than a straight angle. 

If the term polygon is used, a convex polygon is meant. 



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POLYGONS. ' 67 

SOOt Two polygons are equal when they can be divided by 
diagonals into the same number of triangles, equal each to 
each, and similarly placed ; for the polygons can be applied 
to each other, and the corresponding triangles will evidently 
coincide. 

201. Two polygons are mutually equianguh/r^ if the angles 
of the one are equal to the angles of the other, each to each, 
when taken in the same order. Figs. 1 and 2. 

202. The equal angles in mutually equiangular polygons 
are called komohgous angles ; and the sides which lie between 
equal angles are called homologous sides. 

203. Two polygons are mutually equilateral^ if the sides of 
the one are equal to the sides of the other, each to each, when 
taken in the same order. Figs. 1 and 2. 

Fio. 4. Pio. 5. Pio. 6. FiQ. 7. 

Two polygons may be mutually equiangular without being 
mutually equilateral ; as, Figs. 4 and 5. 

And, except in the case of triangles, two polygons may be 
mutually equilateral without being mutually equiangular ; as. 
Figs. 6 and 7. 

If two polygons are mutually equilateral and equiangular, 
they are equal, for they may be applied the one to the other 
80 as to coincide. 

204i A polygon of three sides is called a trigon or triangle; 
one of four sides, a tetragon or quadrilateral; one of five sides, 
a, pentagon; one of six sides, a hexagon; one of seven sides, a 
heptagon; one of eight sides, an octagon; one of ten sides, a 
decagon; one of twelve sides, a dodecagon. 



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68 



PLANE GEOMETRY. 



•BOOK I. 



Proposition XLIV. Theorem. 

305. The sum of the interior angles of a polygon is 
equal to two right angles, taken as many times less 
two as the figure Juis sides. 




Let the figure ABODE F be sl" polygon having n sides. 

To prove ZA+ZB+ZC, etc.=(7i-2) 2 rt.A. 

Proof. From the vertex A draw the diagonals ACy AD^ 
and AU. 

The sum of the A of the A = the sum of the A of the 
polygon. 

Now there are (n — 2) A, 

and the sum of the A of each A = 2 rt. /4. § 138 

.*. the sum of the A of the A, that is, the sum of the A of 
the polygon = {n—2)2rtA, q. e. a 

206, Cor. The sum of the angles of a quadrilateral equals 
two right angles taken (4 — 2) times, i.e., equals 4 right angles; 
and if the angles are all equ^l, each angle is a right angle. In 
general, each angle of an equiangular polygon of n »ide$ is 

equal to y^^ ^ right angles, 
n 



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POLYGONS. 69 



Proposition XLV. Theorem. 

207. The exterior angles of a polygon, made by pro- 
ducing each of its sides in succession, are together 
equal to four right angles. 




Lei the figure ABCDE be a polygon, having its sides 
produced in succession. 

To prove the sum of the ext. A = ^rt. A, 
Proof. Denote the int. A of the polygon by -4, -B, O, D, E, 
and the ext. A by a, 5, c, d, e. 

ZA + Za=^2rtA, §90 

and ZB + Zb^2rt.A, 

(being sup.-adj. A). 

> 

In like manner each pair of adj. ^i = 2 rt. A, 

/. the sum of the interior and exterior zi= 2 rt. ^ taken 
as many times as the figure has sides, 

or, 2 n rt. A. 

But the interior A = 2rt, A taken as many times as the 
figure has sides less two, = (n— 2) 2 rt. -4, 

or, 2 n rt. zS — 4 rt. A. 

/. the exterior -4 = 4 rt. -4. 

€1.1.0. 



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70 



PLANE aEOMETRY. — BOOK I. 



Proposition XL VI. Theorem. 

208. A quadrilateral which has two Ojdj accent sides 
equal, and the other two sides equal, is symmetrical 
with respect to the diagonal joining the vertices of 
the angles formed by the equal sides, and the diagO' 
nals intersect at right angles. 




Let ABCD be a quadrilateral, having 'AB = AD, and 
CB-CD, and having the diagonals AC and BD. 

To prove that the diagonal AC is an axis of symmetry, and 
is J^ to the diagonal BD, 

Proof. Inthe A ^^Can(i^i)(7 

AB = AD, and B0= DO, Hyp. 

and AC=Aa Iden. 

,\AABC=-AADO, * §160 

{Juiving three sides of the one equal to three sides of the other). 

/. Z BAC= A DAC, and Z BCA=Z DCA, 
{homologous A of equal ^). 

Hence, if ABO is turned on AO ^ an axis, AB will fall 
upon AD, OB on OD, and OB on OD, 

Hence ^(7 is an axis of symmetry, § 65. and is ± to BD. 



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POLYGONS. 



71 



Proposition XL VII. Theorem. 

209. If a figure is sj/mmetricaZ with respect to two 
axes perpendicular to ea^h other, it is symmetrical 
with respect to their intersection as a centre. 

T 




Now 



But 



Let the figure ABCDEFGH be symmetrica,! with 
respect to the two axes XX\ YY\ which intersect at 
right angles at O, 

To prove the centre of symmetry of the figure. 

Proof. Let N be any point in the perimeter of the figure. 

Draw NMIl. to FF' and IKLl. to XX\ 

Join LO, ON, and KM. 

KI= KL, 
(the figure being symmetrical with respect to XX'), 

KI= OM, 

(lis comprehended between Ws are eqiuiT). 

/. KL=OM,B,nd KLOMis a O, 

(having two sides equal and parallel). 

.•. LO is equal a^nd parallel to KM. 

In like manner we may prove OJV equal and parallel to KM. 

Hence the points X, 0, and N are in the same straight line 

drawn through the point II to KM\ and LO=ON^ since 

each is equal to KM. 

.'. any straight line LON, drawn through 0, is bisected at 0. 

•*. is the centre of symmetry of the figure. § 64 

Q.E.D. 



§61 
§180 

§182 
§179 



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72 PLANE GEOMETRY. — BOOK! I. 

EXEBCISES. 

^ 34. The median from the vertex to the hase of an isosceles triangle is 
perpendicolar to the base, and bisects the vertical angle. 

35. State and prove the converse. 

^ 36. The bisector of an exterior angle of an isosceles triangle, formed 
by producing one of the legs through the vertex, is parallel to the base. 

37. State and prove the converse. 
' 38. The altitudes upon the legs of an isosceles triangle are equal. 

39. State and prove the converse. 
^40. The medians drawn to the legs of an isosceles triangle are equal. 

41. State and prove the converse. (See Ex. 33.) 

42. The bisectors of the base angles of an isosceles triangle are equal. 

- - 43. State the converse and the opposite theorems. 

44f The perpendiculars dropped from the middle point of the base of 
an isosceles triangle upon the legs are equal ^ 

45. State and prove the converse. 

^ 46. If one of the legs of an isosceles triangle is produced through the 
vertex by its own length, the line joining the end of the leg produced to 
the nearer end of the base is perpendicular to the base. 

T 47. Show that the sum of the interior angles of a hexagon is equal ^ 
eight right angles. 

48. Show that each angle of an equiangular pentagon is { of a right 
angle. 

49. How many sides has an equiangular polygon, four of whose angles 
are together equal to seven right angles ? 

50. How many sides has a polygon, the sum of whose interior angles 
is equal to the sum of its exterior angles ? 

61. How many sides has a polygon, the sum of whose interior angles 
is double that of its exterior angles ? 

52. How many sides has a polygon, the sum of whose exterior angles 
is double that of its interior angles ? 



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EXERCISES. 73 

63. BAG is a triangle haying the angle B douhle the angle A. If BD 
bisect the angle B^ and meet AC in D^ show that BD is equal to AD, 

54. If from any point in the base of an isosceles triangle parallels to 
the legs are drawn, show that a parallelogram is formed whose perimeter 
is constant, and equal to the stun of the legs of the triangle. 

55. The lines joining the middle points of the sides of a triangle divide 
the triangle into four equal triangles. 

56. The lines joining the middle points of the side of a square, taken 
in order, enclose a square. 

57. The lines joining the middle points of the sides of a rectangle, 
taken in order, enclose a rhombus. 

58. The lines joining the middle points of the sides of a rhombus, 
taken in order, enclose a rectangle. 

59. The lines joining the middle points of the sides of an isosceles 
trapezoid, taken in order, enclose a rhombus or a square. 

60. The lines joining the middle points of the sides of any quadri- 
lateral, taken in order, enclose a parallelogram. 

61. The median of a trapezoid passes through the middle points of 
the two diagonals. 

62. The line joining the middle points of the diagonals of a trapezoid 
is equal to half the diflference of the bases. 

63. In an isosceles trapezoid each base makes ^r J> 

equal angles with the legs. / \ \ 

Hint. Draw C^ll DB. / \ \ 

64. In an isosceles trapezoid the opposite angles -^ ^ -S 
are supplementary. 

65. If the angles at the base of a trapezoid are equal, the other 
angles are equal, and the trapezoid is isosceles. 

66. The diagonals of an isosceles trapezoid are equal. 

67. If the diagonals of a trapezoid are equal, the c D 

trapezoid is isosceles. I^\.y\ 

Hint. Draw CE and DF ± to CD. Show that ^ / '•/^\\ \ 
ADF and BCE are equal, that k COD and AOB are //K j\\ 

isosceles, and t}iat ^ AOC andi BOD are equal. 4 ^ p ^ 



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74 ' PLANE GEOMETRY. — BOOK I. 

^ 68. ABOD is a parallelogram, E and F the middle points of AD and 
-5(7 respectively : show that ^j^and DJ^will trisect the diagonal AC. 

69. If from the diagonal BD of a Bqnate ABCD, BE is cnt off equal 
to BC, and EF is drawn perpendicular to BD to meet DC at F^ show 
that DE is equal to EF, and also to FC. 

70. The bisector of the vertical angle ul of a triangle ABC, and the 
bisectors of the exterior angles at the base formed by producing the sides 
AB and AC, meet in a point which is equidistant from the base and the 
sides produced. 

71. If the two angles at the base of a triangle are bisected, and 
through the point of meeting of the bisectors a line is drawn parallel to 
the base, the length of this parallel between the sides is equal to the sum 
of the segments of the sides between the parallel and the base. 

72. If one of the acute angles of a right triangle is double the other, 
the hypotenuse is double the shortest side. 

73. The sum of the perpendiculars dropped from any point in the 
base of an isosceles triangle to the legs is constant, 
and equal to the altitude upon one of the legs. 

Hint. Let PD and PE be the two Js, BF the WjC 

altitude upon AC, ,Draw PG JL to BF, and prove 
the ii PBG and PBD equal. ^ TVy/^^^-^^ n 

^ F 

74. The sum of the perpendiculars dropped from any point within an 
equilateral triangle to the three sides is constant, and equal to the 
altitude. 

Hint. Draw through the point a line il to the base, and apply Ex. 73. 

75. What is the locus of all points equidistant from a pair of inter- 
•ecting lines ? 

76. In the triangt^ CAB the bisector of the angle C makes with the 
perpendicular from Cto AB an angle equal to half the difference of the 
angles A and B. 

77. If one angle of an isosceles triangle is equal to 60^, the triangle 
is equilateral. 




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BOOK IL 
THE CIRCLE. 



Definitions. 

210i A circle is a portion of a plane bounded by a curved 
line called a circumference, all points of which are equally dis- 
tant from a point within called the centre, 

211. A radiics is a straight line drawn from the centre to the 
circumference ; and a diameter is a straight line drawn through 
the centre, having its extremities in the circumference. 

By the definition of a circle, all its radii are equal. All its 
diameters are equal, since the diameter is equal to two radii. 

212. A secard is a straight line which intersects the circum- 
ference in two points ; as, AD, Fig. 1. 

213. A tangent is a straight line which touches the circum- 
ference but does not intersect it; as, 

BC, Fig. 1. The point in which the 
tangent touches the circumference is 
called the point of contact, or jpoint of 
iangency, 

214. Two circumferenceB are tangent 
to each other when they are both tan- ^^^' ^* 

gent to a straight line at the same point; and are tangent 
internally or externally, according as one circumference lies 
wholly within or without the other. 




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76 



PLANE GEOMETRY. 



-BOOK II. 



215. An arc of a circle is any portion of the circumference. 
An, arc equal to one-half the circumference is called a semi- 
drcu^nfetence, 

216. A chord is a straight line having its extremities in the 
circumference. 

Every c^hord subtends two arcs whose sum is the circum- 
ference ; thus, the chord AB (Fig. 3) subtends the smaller arc 
AB and the larger arc BCDEA, If a chord and its arc are 
spoken of, the less arc is meant unless it is otherwise stated. 






217. A segment of a circle is a portion of a circle bounded 
by an arc and its chord. 

A segment equal to one-half the circle is called a semicircle, 

218. A sector of a circle is a portion of the circle bounded 
by two radii and the arc which they intercept. 

A sector equal to one-fourth of the circle is called a quadrant. 

219. A straight line is inscribed in a circle if it is a chord. 

220. An angle is inscribed in a circle if its vertex is in the 
circumference and its sides are chords. 

221. An angle is inscribed in a segment if its vertex is ou 
the arc of the segment and its sides pass through the extrem- 
ities of the arc. 

222. A polygon is inscribed in a circle if its sides are 
chords of the circle. 

288. A circle is inscribed in a polygon if the circumference 
Ruches the sides of the polygon but does Qot; ipt^r^^ct tbe®. 



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ARCS AND CHORDS. 77 

A polygon is circumscribed about a circle if all the 
sides of the polygon are tangents to the circle. 

225. A circle is circumscribed about a polygon if the circum- 
ference passes through all the vertices of the polygon. 

226. Two circles are equal if they have equal radii ; for 
they will coincide if one is applied to the other ; conversely, 
two equal circles have equal radii. 

Two circles are concentric if they have the same centre. 

Proposition I. Theorem. 

227. The diameter of a circle is greater than any 
other chord; and bisects the circle and the circum' 
ferenoei, 




p 
Let AB be the diameter of the circle AMBP, and 
AE any other chord. 

To prove AB > AE, and AB bisects the circle and the 
circumfereTice, 
2toot I. From C, the centre of the O, draw CH. 
CU=-CB, 
(being radii of the same circle). 

But AO+CE>AE, §137 

(the sum of two sides of a A is ">> the third side). 

Then A0+ OS > AE, or AS > AE. Ax. 9 

II. Fold over the segment AMB on AB as an axis until it 
falls upon APB, § 59. The points A and B will remain fixed; 
therefore the arc AMB will coincide with the arc APB\ 
because all points in each are equally distant from the 
centre C. § 210 

Hence the two figures coincide throughout and are equal. § 59 

Q.B.a 



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78 PLANE GEOMETRY. — BOOK 11. 



Proposition II. Theorem. 

228. A straight line cannot intersect the circitfnr 
ference of a circle in more than two points. 




LetHKbe anjr line cutting the circumference AMP. 

To prove that HK can intersect the circumference in only two 
points. 

Proof. If possible, let HK intersect the circumference in 
three points JT, P, and K. 

From 0, the centre of the O, draw O-ff, OP, and OK 

Then OK, OF, and OJTare equal, 

(being radii of the aame circle). 

Hence, we have three equal straight lines OK, OP, and OK 
drawn from the same point to a given straight line. But this 
is impossible, § 120 

{only ttoo equal straight lines can he drawn from a point to a straight line). 

Therefore, KK can intersect the circumference in only two 
points. a c a 



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ARCS AND CHORDS. 79 



. Proposition III. Theorem. 

229. In the same circle, or equal circles, equal an^ 
gles at the centre intercept equal arcs; conversely, 
equal arcs subtend eqiuxl angles at the centre. 





p p' 

In the equal circles ABP and AfB'P' let Z0m:Z0f. 

lb prove arc BS= arc i?'/S". 

Proot Apply O ABP to O A'JB'P, 

so that Z shall coincide with Z O, 

R will fall upon R\ and 8 upon 8\ § 226 

(Jor OR^ 0'E\ and 08 => 0^8^, being radii of equal ©). 

Then the arc US will coincide with the arc i?'/S", 

since all points in the arcs are equidistant from the centre. 

§210 
.•.arc^/S=arc R'S^. 

CoNVBBSELr : Let arc RS = arc R'S', 

Toprove ZO = Z(y. 

Proof. Apply O ABP to O A'B'P', so that arc BS shall fall 
upon arc B'S^ B falling upon B\ S upon /?, and upon O*. 

Then BO will coincide with ^'0', and 80 with /S'O'. 

,\^0 and 0' coincide and are equal. q. e. ^ 



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80 PLANE GEOMETRY. — BOOK II. 

Proposition IV. Theorem. 

230. In the same circle, or equal circles, if two 
chords are equal, the arcs which they subtend are 
equal; conversely, if two arcs are equal, the chords 
which subtend them are equal. 





p p' 

JjB the equal circles ABP and A'B'Pf, let chord RS = 
chord R'Sf. 

To prove arc BS = arc B^S^. 

Proof. Draw the radii OB, OS, aR\ and CK/S". 

In the A 0R8 and O'^'/S' 

R8=B8\ Hyp. 

the radii OR and 08= the radii ON and 08*, § 226 

.\AR08=ARa8\ §160 

ifhru sides of the one being equal to three sides oj the other). 

.\ZO = Z&, 

.-. arc R8= arc R'8', § 229 

(in equal ®, equal A at the centre intercut equal arcs). 

CoNVEESELY ; Let ATc RS = arc R^JS', 
To prove chord R8 = chord R*8'. 

Proof. ZO = ZO\ §229 

(equal arcs in equal (D subtend equal A at the centre), 
and OR and 08= OR* and a8\ respectively. § 226 

.\A0R8=AaR'8\ §160 

ijiaving two sides equal each to each and the included A equal). 

.-. chord R8= chord R'8\ ^^^ 



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ARCS AND CHORDS. 81 



Proposition V. Theorem. 

231. In the same circle, or equal circles, if two arcs 
are unequal, and each is less than a semi^circumfer" 
ence, the greojter arc is subtended by the greater 
chard; conversely, the greater chord subtends the 
greoiter arc. 




Xa the circle whose centre is 0, let the Arc AMB be 
greater than the arc AMF, 

To prove chxyrd AB greater than chord AF. 
Proof. Draw the radii OA, OF, and OB. 

Since F'ls between A and B, OF will fall between OA and 
OB, and Z AOB be greater than A AOF. 
Hence, in the h^ AOB and AOF, 

the radii OA and OB = the radii OA and OF, 
but Z AOB is greater than Z AOF. 

:. AB > AF, § 152 

(^ iL having two sides equal each to each, but the included A unequ4il). 

OoHYEBSELT : Let AB bc greater than AF. 
To prove are AB greater than arc AF, 
In the A AOB and AOF, 

OA and 0B= OA and OJP respectively. 
But AB is greater than AF. Hyp, 

.-. Z AOB is greater than Z AOF, ^_^ § i58 
Ifl^ J^ having tuH> ddea equal each to each, hui the third eiM unequal). 
.-. OB falls without OF 
/. arc AB is greater than arc AF a e. ex 



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82 PLANE GEOMETRY. — BOOK II. 



Proposition VI. Theorem. 

233. The radius perpendicular to a chord bisects 
the chord and the arc subtended by it. 

E 




S 

Let AB he the chord, and let the radius OS be per' 
pendicnlar to AB at M. 

To prove AM= BM, and arc A8 = arc £8. 

Pvoof. Draw OA and 0£ from O, the centre of the circle. 

In the rt. A 0AM md OBM 

the radius OA = the radius OB^ 

and 0M== OM. Iden. 

.'.AOAM=AOBM, §161 

(hamng the hypotenuse and a nde of one equal to the hypotenuu and a 
tide of the other). 

.'.AM=BM, 
BjidZA08=j^B08. 
.-. arc A8= arc B8, § 229 

(equal Aatihe eevUre intercept eqybdl arcs on the cirewnferenu), 

aE.0. 
283. Cor. 1. The perpendicular erected at the middle of a 
chord passes through the centre of the circle. For the centre is 
equidistant from the extremities of a chord, and is therefore in 
the perpendicular erected at the middle of the chord. § 122 
234. Cor. 2. The perpendicular erected at the middle of a 
chord bisects the arcs of the chord, 

236. Cor. 8. The locus of the middle poinis of a sj/stem of 
parallel chords is the diameter perpendicular to them. 



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ARCS AND CHORDS. 83 



Proposition VII. Theorem. 

236. In the same circle, or equal circles, equal 
chords are equally distant from the centre; and 

GONYEBSBLY. 




Let AB and CF be equal chords of the circle ABFC. 

To prove AB and CF equidistarU from the centre 0. 

Proof. Draw OPl. to AB, OHl. to CF, and join OA and OC. 

OP and OH bisect AB and OF, § 232 

{a radius ±to a chord bisects it)* 

Hence, in the rt. A OF A and OHC 

af=cjb:, Ai. 7 

the radius OA = the radius 00. 

,',AOFA = AOIia §161 

(hanng a tide and hypotenuse of the one equal to a side and hypotenuse^ 
of the other). 

:,0P=OH. 
.\ AB and CFare equidistant from 0. 

COFYEBSELY : Let OP = OH, 

To prove AB=CF. 

Pioof. In the rt. A OP-4 and OJTC 
the radius OA = the radius OC, and OP^ OH. Hyp. 

/. A OP A and OHC are equal. ^ § 161 

/. AP^ OS. 

,\AS^OF. Ax 6. 



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84 PLANE GBOMETRY. — BOOK II. 



Pboposition VIII. Theorem. 

237t In the same circle, or equal circles, if two 
chords are unequal, they are unequally distant from 
the. centre, and the greater is at the less distance. 




In the circle whose centre is 0, let the chords AB 
and CD be unequal, and AB the greater; and let OE 
and OF be perpendicular to AB and CD respectively. 

To prove OE < OF, 

fiwt. Suppose AO drawn equal to CD, and OSl, to AO. 

Then 0H= OF, § 236 

(jm the tame Q two equal chords are equidutarU from the centre). 

Join EH, 

OE and OJT bisect AB and AO, respectively, § 232 
(a radiui Xtoa chord Useeti it). 

Since, by hypothesis, AB is greater than CD or its equal A O, 
AE, the half of AB, is greater than AH, the half oi AO. 

.'. the Z AHE is greater than the Z AEH, § 158 
{the greater of two sides of a A has the greater Z opposite to it). 

Therefore, the Z OHE, the complement of the Z AHE, is 
less than the Z OEH, the complement of the Z AEH, 

.\OE<OH, §159 

{the greater of two A of a A Tias the greater side opposite to it), 

' .-. OE < OF, the equal of OH 



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ARCS AND CHORDS. 85 



Proposition IX. Theorem. 

238. Conversely : In the same circle, or equal cir- 
cles, if two chords are unequally distant from the 
centre, they are unequal, and the chord at the less 
distance is the grealer. 




In the circle whose centre is 0, let AB and CD be 
unequally distant from 0; and let OE perpendicular 
to AB he less than OF perpendicular to CD. 

To prove AB>CD. 

Proof. Suppose AO drawn equal to CD, and OSl. to AQ. 

Then OR^ OF, §236 

{in the same O two equal chorda are equidistant from t?ie centre). 

Hence, OE < OH. 

Join Uff. 

In the A OERthe Z OHE is less than the Z. OEH, % 158 

{),he greater of two sides of a A has the greater Z opposite to it). 

Therefore, the Z AHE, the complement of the Z OHE, is 
greater than the Z AEH, the complement of the Z OEH. 

.\AE>AH, §159 

{flk€ greater of two A of a A has the greater side opposite to %t). 

But AE-=^\AB, and AH=\AO. 

^ .•. AB > AO; hence AB > CD, the equal of AO. 

aE.a 



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86 PLANE GEOMETBT. —BOOK n. 



Pboposition X. Theorem. 

839. A straight line perpendicular to a radius at 
Us extremity is a tangent to the circle. 




H A 

Let MB be perpendicular to the radius OA at A 

To prove MB tangent to the circle. 

Proof. From draw any other line to MB, as OCfl". 

OH>OA, §114 

{aXUihe ahortest line from a point to a straight Kne). 

.*. the point J?" is without the circle. 
Hence, every point, except A, of the line MB is without the 
circle, and therefore MB is a tangent to the circle' at A. § 213 

aB.0. 

240. Cor. 1. A tangent to a circle is perpendicular to the 
radius drawn to the point of contact. For, if MB is tangent 
to the circle at A, every point of MB, except -4, is without 
the circle. Hence, OA is the shortest line from to MB, and 
is therefore perpendicular to MB (§ 114) ; that is, MB is per- 
pendicular to OA. 

241. CoR. 2. A perpendicular to a tangent at the point of 
contact passes through the centre of the circle. For a radius is 
perpendicular to a tangent at the point of contact, and there- 
fore, by § 89, a perpendicular erected at the point of contact 
coincides with this radius and passes through the centre. 

242. OoR. 3. A perpendicular let fall from the centre of a 
circle upon a tangent to the circle passes through the point of 
contact 



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ARCS AND CHORDS. 



87 



Proposition XI. Theorem. 

243. Parallels intercept equal arcs an a circumr- 
ference. 






Fig. 1. 



§241 
§102 
§232 



Fig. 2. 
Let AB and CD be the two parallels. 

Case I. When AB is a tangent, and CD a secant. 

Suppose AJB touches the circle at K 
To prove arc CF= arc DF. 

Proof. Suppose FF' drawn ± to -45. 

This X to AB at -F is a diameter of the circle. 

It is also ± to CD, 

.\ arc CF= arc DF, 

(a radius Xto a chord bisecU the chord and its subtended arc). 

Also, arc FCF' = arc FDF\ § 227 

,',aTc(FCF-FC)^B.rc(FDF-FD), § 82 
that is, arc CF' = arc DF, 

Case II. When AB and CD are secants. Fig. 2. 
Suppose -EF drawn II to CD and tangent to the circle at M. 

Then arc AM = arc BM 

and arc CM = arc D3f- Case I. 

/. by subtraction, arc AC = arc BD. 

Case III. When AB and CD are tangents. Fig. 3. 
Suppose AB tangent at F, CD at F, and GH II to AB. 

Then arc OF = arc FE Case I. 

and arc OF — arc HF, 

/. by addition, arc FOF= arc EHF. a t a 



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88 PLANE GEOMETRY. — BOOK II. 

Proposition XII. Theorem. 

244. Through three points not in a straight line, 
one oircumferenoe, and only one, can be drawn. 



Let A, B, C be three points not in a straight line. 

To prove that a circumference can be drawn through -4, -B, 
and C, and only one. 

Proof. Join AB and £0. 

At the middle points of AB and -BC suppose Js erected. 

Since BO is not the prolongation of -4-B, these Js will inter- 
sect in some point 0. 

The point 0, being in the ± to AB at its middle point, is 

equidistant from A and -S ; and being in the J- to BO at its 

middle point, is equidistant from B and (7, § 122 

{every point in the perpendicular 'bisector of a straight line is equidistant 

from the extremities of the straight line). 

Therefore is equidistant from -4, B, and 0; and a cir- 
cumference described from as a centre, with a radius 0-4, 
will pass through the three given points. 

Only one circumference can be made to pass through 
these points. For the centre of a circumference passing 
through the three points must be in both perpendiculars, and 
hence at their intersection. As two straight lines can inter- 
sect in only one point, is the centre of the only circumfer- 
ence that can pass through the three given points. q. e. d. 
^245. Cor. Two circumferences can intersect in only two 
points. For, if two circumferences have three points common, 
they coincide and form one circumference. 



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TANGENTa 89 

Proposition XIII. Theorem. 

246. The tangents to a circle drawn from an exte- 
rior point are equal, and make equal angles with 
the line Joining the point to th^ centre. 




C 

Let AB and AC be tangents from A to the circle 
whose centre is O, and AO the line Joining A to 0. 

To pr(yve AB = AO, and Z BAO = Z CAO. 
Proof. Draw OB and OC. 

AB is ± to OB, and AC ± to OC, § 240 

(a tarigmt to a circle is ± to the radiu% dravm to the point of contact). 
In the rt. A OAB and OAC 

OB^OO, 
{yadii of the same circle), 

OA = OA, Iden. 

.\l^OAB = AOAC, §161 

{having u side and hyjpotknuse of the one equal to a side and hypotenuse 
of thz othfir). 

.\AB^AC, 

and Z BAO = Z CAO. a e. d. 

247. Dep. The line joining the centres of two -circles is 
called the line of centres. 

248. Def. a common tangent to two circles is called a 
common exterior tangent when it does not cut the line of cen- 
tres; and a common interior tangent when it cuts the line of 
centres. 



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90 PLANE GEOMETBY. — BOOK II, 



Peoposition XIV. Theorem. 

248. If two circumferences intersect each other, the 
line of centres is perpendicular to their common 
chord at its middle poitjjb^ T \ 




Let and O he the centres of two circumferences 
which intersect at A and B. Let AB be their common 
chord, and CC Join their centres. 

To prove CO* ± to AB at its middle point. 

Proof. A ± drawn through the middle of the chord AB 
passes through the centres C and 0", § 233 

(a ± erected at the middle of a chord passes through the centre of the Q). 

.*. the line CC\ having two points in common with thisX, 
must coincide with it. 

.'. CC" is X to AB at its middle point. q. e. d. 



Ex. 78. Describe the relative position of two circles if the line of 
centres : \ 

(i.) is greater than the sum of the radii ; 
(ii.) is equal to the sum of the radii ; 

(iii.) is less than the sum but greater than the difference of th« ni^ ; 
(iv.) is equal to the difference of the radii ; 
(y.) is less than the difference of the radii. 
Illustrate each case by a figure. 



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TANGENTS. 91 



Proposition XV. Theorem. 

350. If two circumferences are tangent to ea/ih other, 
the line of centres -passes through the -point of contact. 




Let the two circumferences, whose centres are C 
and C, touch each other at O, in the straight line AB, 
and let CO be the straight line joining their centres. 

To prove Oisin the straight line 00\ 

Proof. A J- to AB, drawn through the point 0, passes 
through the centres C and C\ § 241 

{p, l^ioa tangent at the point of contact passes through the centre 
of the circle). 

.-. the line CO', having two points in common with this ± 
must coincide with it. 



is in the straight line CO'. 



Q.E.O. 



Ex. 79. The line joining the centre of a circle to the middle of a 
chord is perpendicular to the chord. 

Ex. 80. The tangents drawn through the extremities of a diameter 
ftre parallel. 

Ex. 81. The perimeter of an inscribed equilateral triangle is equal 
to half the perimeter of the circumscribed equilateral triangle. 

Ex. 82. The sum of two opposite aides of a circumscribed quadri- 
lateral is equal to the sum of the other two sides. 



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92 PLANE GEOMETRY. — BOOK IL 



Measurement. 

26L To measure a quantity of any kind is to find how many 
times it contains another known quantity of the same kind. 

Thus, to measure a line is to find how many times it con- 
tains another known line, called the linear unit. 

The number which expresses how many times a quantity 
contains the unit-quantity, is called the numerical measua^e 
of that quantity ; as, 5 in 5 yards. 

252. The magnitude of a quantity is always relative to the 
magnitude of another quantity of the same kind. No quantity 
is great or small except by comparison. This relative magni- 
tude is called their ratio^ and is expressed by the indicated 
quotient of their numerical measures when the same unit of 
measure is applied to both. 

The ratio of a to b is written f , or a : b, 

b 

253. Two quantities that can be expressed in integers in 
terms of a common unit are said to be commensurable. The 
common unit is called a common measurCy and each quantity 
is called a multiple of this common measure. 

Thus, a common measure of 2^ feet and 8f feet is •)■ of a 
foot, which is contained 15 times in 2^ feet, and 22 times in 
3f feet. Hence, 2^ feet and Sf feet are multiples of ^ of a 
foot, 2^ feet being obtained by taking ^ of a foot 15 times, and 
3f feet by taking J of a foot 22 times. 

264. When two quantities are incommensurable, that is, 
have no common unit in terms of which both quantities can be 
expressed in integers, it is impossible to find a fraction that 
will indicate the exact value of the ratio of the given quanti- 
ties. It is possible, however, by taking the unit sufficiently 
small, to find a fraction that shall differ from the true value 
of the ratio by as little as we please. 



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RATIO. 93 

^ Thus, suppose a and b to denote two. lines, such that 

a a /- 

b J = ^- 

Now V2= 1.41421356 , a value greater than 1.414213, 

but less than 1.414214. 

If, then, a millionth part of b be taken as the unit, the value 

of the ratio f lies between |^^^^j and |HHM» *^<i there- 
o 

fore differs from either of these fractions by less than unrimnr- 
By carrying the decimal farther, a fraction may be found 
that will differ from the true value of the ratio by less than a 
billionth, a trillionth, or any other assigned value whatever. 

Expressed generally, when a and b are incommensurable, 
and b is divided into any integral number (n) of equal parts, 
if one of these parts is contained in a more than m times, but 
less than m + 1 times, then 

^>— , but < — J--; 
b n n 

that is, the value of % lies between — and ^— L_. 
b n n 

The error, therefore, in taking either of these values for 

- is less than -. But by increasing n indefinitely , - can be 
on n 

made to decrease indefinitely^ and to become less than any 
assigned value, however small, though it cannot be made 
absolutely equal to zero. 

Hence, the ratio of two incommensurable quantities cannot 
be expressed exactly by figures, but it may be expressed ap- 
proximately within any assigned measure of precision. 

266. The ratio of two incommensurable quantities is called 
an incommensurable ratio ; and is a fixed value toward which 
its successive approximate values constantly tend. 



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94 PLANE GEOMETRY. — BOOK II. 

25& Theobem. Ihvo incommensurable ratios are equal if, 

when the unit of measure is indefinitely diminished^ their ap- 
proximate values constantly remain equ^al. 

Let a:b and a' : i' be two incommensurable ratios whose true 

values lie between the approximate values — and ^ ' , 

n n 

when the unit of measure is indefinitely diminished. Then 

they cannot differ so much as -• 

n 

Now the difference (if any) between the fixed values a : b 

and a' : ^', is & fixed valv^. Let d denote this difference. 

Then rf<i. 

n 

But if d has any value, however small, -, which by hypoth- 

n 

esis can be indefinitely diminished, can be made less than d. 

Therefore d cannot have any value; that is, c? = 0, and 
there is no difference between the ratios a : b and a' : 6'; there- 
fore a:b = a' :bK 

The Theoby op Limits. 

257. When a quantity is regarded as having o, fixed Ya\\xe 
throughout the same discussion, it is called a constant ; but 
when it is regarded, under the conditions imposed upon it, as 
having different successive values, it is called a variable. 

When it can be shown that the value of a variable, measured 
at a series of definite intervals, can by continuing the series 
be made to differ from a given constant by less than any 
assigned quantity, however small, but cannot be made abso- 
lutely equal to the constant, that constant is called the Umii 
of the variable, and the variable is said to approach indefir 
nitely to its limit. 

If the variable is increasing, its limit is called a supericf 
limit ; if decreasing, an inferior limit. 



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THEORY OF LIMITS. 95 

Suppose a point to move from A toward -B, under the con- 
ditions that the first ^ m jt w b 



second it shall move ' ' * 

one-half the distance from ^ to ^, that is, to M\ the next 
second, one-half the remaining distance, that is, to W\ the 
next second, one-half the remaining distance, that is, to Jf " ; 
and so on indefinitely. 

Then it is evident that the moving point 'may approach as 
near to £ as we please^ but mil never arrive at B, For, how- 
ever near it may be to £ at any instant, the next second it 
will pass over one-half the interval still remaining ; it must, 
therefore, approach nearer to -B, since half the interval still 
remaining is some distance, but will not reach B, since half 
the interval still remaining is not the whole distance. 

Hence, the distance from A to the moving point is an in- 
creasing variable, which indefinitely approaches the constant 
AB as its limit ; and the distance from the moving point to 
jB is a decreasing variable, which indefinitely approaches the 
constant zero as its limit. 

If the length of AB is two inches, and the variable is 
denoted by ar, and the difference between the variable and its 
limit, by v : 

after one second, ar = 1, v = 1 

after two seconds, ar=l-f-^, v = i 

after three seconds, x=l + ^ + \j v = \ 
after four seconds, ar = l + J + i + ^, v = i 
and so on indefinitely. 

Now the sum of the series 1 + ^ + i + i» etc., is less than 
2 ; but by taking a great number of terms, the sum can be 
made to differ from 2 by as little as we please. Hence 2 is 
the limit of the sum of the series, when the number of the 
terms is increased indefinitely ; and is the limit of the dif- 
ference between this variable sum and 2. 



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96 PLANE GEOMETEY. — BOOK II. 
Consider the repetend 0.33333 which may be written 

TS + WIT +10 I 10000 + 

However great the number of terms of this series we take, 
the sum of these terms will be less than -^; but the more 
terms we take the nearer does the sum approach i. Hence 
the sum of the series, as the number of terms is increased, 
approaches indefinitely the constant -^ as a limit. 

258. In the right triangle -4CB, if the vertex A approaches 
indefinitely the base -B(7, the angle £ a 
diminishes, and approaches zero indefi- 
nitely ; if the vertex A moves away from 
the base indefinitely, the angle £ increases 
and approaches a right angle indefinitely ; 
but £ cannot become zero or a right angle, 
so long as AC£ is a triangle ; for if B be- ^^ 
comes zero, the triangle becomes the straight line BO, and if 
B becomes a right angle, the triangle becomes two parallel 
lines AC&ni AB perpendicular to BC, . Hence the value of 
B must lie between 0** and 90® as limits. 

259. Again, suppose a square A BCD inscribed in a circle, 
and E, F, H, K the middle points of the arcs subtended by 
the sides of the square. If we draw 
the straight lines AE, EB, BF, etc., 
we shall have an inscribed polygon of 
double the number of sides of the 




square. 

The length of the perimeter of this 
polygon, represented by the dotted 
lines, is greater than that of the 
square, since two sides replace each 
side of the square and form with it a triangle, and two sides 
of a triangle are together greater than the third side ; but less 
than the length of the circumference, for it is made up of 




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THEORY OP LIMITS. 97 

straight lines, each one of which is less than the part of the 
circumference between its extremities. 

By continually repeating the process of doubling the num- 
ber of sides of each resulting inscribed figure, the length of 
the perimeter will increase with the increase of the number 
of sides ; but it cannot become equal to the length of the cir- 
cumference, for the perimeter will continue to be made up of 
straight lines, each one of which is less than the part of the 
circumference between its extremities. 

The length of the circumference is therefore the limit of the 
length of the perimeter as the number of sides of the inscribed 
figure is indefinitely increased. 

280. Theorem. If two variables are constantly equal 
and each approojches a limit, their limits are equal. 

D 




At ^__^_ 

-0 

Let AM and AN be two variables which are con- 
stantly equal and which approach indefinitely AB 
and AC respectively as limits. 

To prove A£ = Aa 

Proof. If possible, suppose AB > AC, and take AD = AC, 

Then the variable AMm&j assume values between AD and 
A£f while tlie variable AN must always be less than AD. 
But this is contrary to the hypothesis that the variables should 
continue equal. 

.*. A£ cannot he > AC. 

In the same way it may be proved that -4(7 cannot be >A£. 

.'■ AB and ^(7 are two values neither of which is greater 
than the other. 

UmcQAB^AO. ^^^ 



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98 



PLANE GEOMETRY. — BOOK IL 



Measure of Angles. 

Proposition XVI. Theorem. 

261. In the same circle, or equal circles, two angles 
aib the centre have the same raMo as their intercepted 
arcs. 





Case I. When the arcs are commensurahle. 

In the circles whose centres are C and D, let ACS and 
BDF he the angles, AB and EF the intercepted arcs* 

Toprove ^ACB^^roAB^ 

^ /.EDF tiTcEF 

Proofi Let m be a common measure of A£ and UF. 

Suppose m to be contained in AB seven times, 
and in FF four times. 
arc AB _ 7 
4 



Then 



(1) 



arc FF 

At the several points of division on AB and FF draw radii. 
These radii will divide ZACB into seven parts, and 
Z -E'Z^jPinto four parts, equal each to each, § 229 

(in the same O, or equal ®, equal arcs whtend equal Jtatthe centre). 

. ZACB __7 
'/.FDF 4 
From (1) and (2), 

ZAOB B.rcAB 



(2) 



Z FVF arc EF 



Ax,l 



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MEASURE OP ANGLES. 99 

Oase II. When the area are ineommenmrable. 
p P' 





In the equal circles ABP and A'B'P* let the angles 
ACB and A'C*^ intercept the incommensurable arcs 
AB and A!B. 

7^^^.. ^ ACB _ arc AB 

Proof. Divide AB into any number of equal parts, and 
apply one of these parts as a unit of measure to A^B^ as many 
times as it will be contained in A^B\ 

Since AB and A^B^ are incommensurable, a certain number 
of these parts will extend from -4' to some point, as i>, leav- 
ing a remainder DB^ less than one of these parts. 

Draw C^D. 

Since AB and A^D are commensurable, 

Z.AOB __ B,rc4B 
Z A^C'D arc A'D 

If the unit of measure is indefinitely diminished, these ratios 
continue equal, and approach indefinitely the limiting ratios 

ZAOB ^^^ B.TCAB 



Z A'C'B' arc A'B* 

Therefore AA^ = arcjLB^ « 23^ 

{Jff tfwo variabUt are eomtanUy equal, and each approaehee a limit, their 
hmU$ are equ>aL) 



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100 



PLANE GEOMETRY. — BOOK II. 



262. The circumference, like the angular magnitude about 
a point, is divided into 360 equal parts, called degrees. The 
arc-degree is subdivided into 60 equal parts, called minutes ; 
and the minute into 60 equal parts, called seconds. 

Since an angle at the centre has the same number of angle- 
degrees, minutes, and seconds as the intercepted arc has of arc- 
degrees, minutes, and seconds, we say : An angle at the centre 
is measu.red by its intercepted arc; meaning. An angle at the 
centre is stick a part of the whole angular magnitude about 
the centre as its inteixepted arc is of the whole circumference. 

Proposition XVII. Theorem. 

263. An inscribed angle is mecbsured by one-half 
the are intercepted between its sides, 

B B B^ 






Case I. When one side of the cmgle is a diameter. 
In the circle PAB (Fig, 1), let the centre C be in 
one of the sides of the inscribed angle B. 

To prove Z,B is measured by ^ arc PA. 
Proof. Draw CA. 

Radius CA = radius OB. 

.\Z£ = ZA, §154 

{being opposite equal tides of the A QAB), 
But ZPOA = ZB + ZA, §145 

{the exterior /.of a Lis equal to the sum of the two opposite interior A). 

.\ZF0A = 2ZB. 

But Z PCA is measured by PA, § 262 

{tJ^e Z at the centre is measured by the intercepted are). 

.'.Z B is measured by J PA. 



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MEASURE X)F j^C^LES:/-': :/ /. : : ;101 

Case II. When the centre is within the angle. 

In the circle BAE (Fig. 2), let the centre C fall 
within the angle EBA. 

To prove Z EJ5A is measured by ^ arc EA, 
Proof. Draw the diameter BCP. 

Z PBA is measured by \ arc PA, Case I. 

Z PBE is measured by ^ arc PEt Case I. 

. • . Z PBA + Z PBE is measured by \ (arc PA + arc PE), 
or Z -£7-5-4 is measured by ^ arc JL4. 
Case III. When the centre is without the angle. 
In the circle BFP (Fig. 3), let the centre C fall 
without the angle ABF, 
To prove Z ABF is measured by ^ arc AF, 
Proof. Draw the diameter BCP. 

Z PBF is measured by \ arc PF, Case I. 

Z PBA is measured by \ arc PA. Case I. 

.'.Z PBF -A PBA is measured by \ (arc PF~ arc P-4), 
or Z ABF is measured by ^ arc -4i^ a e. a 

A^ ^ B^ 




Fig. 1. Pig. 2. Fig. 8. 

264. Cor. 1. An angle inscribed in a semicirck is a right 
angle. For it is measured by one-half a semi-circumference. 

265. Cor. 2. An angle inscribed in a segment greater than a 
semicircle is an acute angle. For it is measured by an arc less 
than half a semi-circumference ; as, Z CAD. Fig. 2. 

266. CoR. 3. An angle inscribed in a segment less than a 
semicircle is an obtuse angle. For it is measured by an arc 
greater than half a semi-circumference ; as, Z CBJD. Fig. 2. 

267. CoR. 4. All angles inscribed in the same segment are 
eqv^al. For they are measured by half the same arc. Fig. 3. 



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102 FLANB QEOMEteY. — BOOK n. 



Peoposition XVIII. Theorem. 

268. An angle formed by two chords intersecting 
within the circumference is measured by one-half 
the sum of the intercepted arcs. 



Let the angle AOC be formed by the chords AB 
and CD. 

To prove Z AOOis measured by ^ {AC+ BD). 
Proot Draw AD. 

ZOOA = ZD + ZA, §145 

(the exterior ZofaAU equal to the sum of the two opposite interior A). 

But ZJDia measured by ^ arc AC, § 263 

and Z A ia measured by i arc £D, 

(an inscribed Z is measured by i the interested arc). 

.'.Z GO A is measured by \ (^C+ BD). 

Q.E.a 



Ex. 83. The opposite angles of an inscribed quadrilateral are sup- 
plements of each other. 

Ex. 84. If through a point within a circle two perpendicular chords 
are drawn, the sum of the opposite arcs which they intercept is equal to 
a semi-circumference. 

Ex. 85. The line joining the centre of the square described upon the 
hypotenuse of a rt. A, to the vertex of the rt. Z, bisects the right angle. 

HiiiT. Describe a circle upon the hypotenuse as diameter. 



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MEASURE OF ANGLES. 103 



Proposition XIX. Theorem. 

269. An angle formed hy a tangent and a chcrd is 
measured by one-half the intercepted arc 




Let MAH be the angle formed by the tangent MO 
and chord AH, 

To prove Z. MAH is measured hy \ arc AEH. 

VvxIL Draw the diameter ACF. 

ZMAFkeLTtZ, §240 

(ihe radius drawn to a tangent at the point of contact UXto it). 

Z MAF being a rt. Z, is measured by \ the semi-circum- 
ference AEF. 

But Z -ELl-Pis measured by J arc JSTF, §263 

(an immbed A is measured hy \ the intercepted arc). 

.-. Z MAF-Z SAFia measured by i{AFF- HF) ; 

or Z MAH\& measured by ^ AEH. 

acD. 

Ex. 86. If two circles touch each other and two secants are drawn 
through the point of contact, the chords joining their extremities are 
parallel. Hint. Draw the common tangent. 



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104 



PLANE GEOMETRY. — BOOK II. 



Proposition XX. Theorem. 

270. An angle formed by two secants, tivo tangents, 
or a tangent and a secant, intersecting without the 
circumference, is measwred by ane^haZf the difference 
of the intercepted arcs. 




FlQ. 1. 




Fia. a 



Case I. Angle formed by two secants. 

Let the angle (Fig. 1) be formed by the two se- 
cants OA and OB, 

To prove /lOis measured by \ (AB — EO), 

Proof. Draw CB. 

ZACB = ZO + ZB, §145 

(the exterior Zofa AU equal to the sum of the two opposite irUerior A). 

"By taking away Z B from both sides, 

ZO = ZAOB-ZB, 
But ZACBis measured by ^ AB, § 263 

and Z B is measured by -J- OE, 

(an inscribed Z. is measured by J the interested are), 

.-. Z O is measured by | (AB— CE), 



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MEASURE OP ANGLES. 106 

Case II. Angle formed hy two tangents. 

Let the angle (Fig. 2) be formed by the two tan- 
gents OA and OB, 

To prove Z is measured by \ {AMB — ASB). 

Proof. Draw AB, 

Z ABC=/, + Z OAB, § 145 

{the exterior /.of a C^ is eqital to the sum of the two opposite interior A). 

By taking away Z OAB from both sides, 

ZO = ZABO-ZOAB. 

But Z ABO is measured by i AMB, § 269 

and Z OAB is measured by ^ ASB, 

(an /formed by a tangent and a chord is measured hy J the intercepted arc). 

/. Z is measured by ^ (AMB— ASB), 

Case III. Angle formed hy a tangent and a secant. 

Let the angle (Fig, 3) be formed by the tangent 
OB and the secant OA, 

To prove Z is measured by ^ {ADS — OES). 
Proof. Draw OS. 

Z AOS=Z O + Z OSO, § 145 

{the exterior /of a IS. is equal to the mm of the two opposite interior A). 

By taking away Z OSO from both sides, 

ZO==ZAOS-ZOSO, 

But ZAOSia measured by | ADS, § 263 

{being an inscribed /), 

and Z OSO is measured by ^ OES, § 269 

(being an /formed by a tangent and a chord). 

,'. ZOis measured by ^{ABS— OES). 

QLE.D, 



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106 



PLANE GEOMETRY. — BOOK II. 



Pboblems of Construction. 

Proposition XXI. Problem. 

271. At a given point in a straight line, to erect a 
perpendicular to that line. 



B 




HOB 

Fia. 1. 

I. Let be the given point in AC. (Fig. 1). 
To erect a Ju to the line A Oat the point 0. 

Oonstmotion. From as a centre, with any radius OB, 
describe an arc intersecting AO in two points J7"and B, 

From J7" and B as centres, with equal radii greater than 
OB, describe two arcs intersecting at -B. Join OB. 

Then the line OB is the X required. 

Proof. Since and B are two points at equal distances from 
-ffand Bf they determine the position of a perpendicular to 
the line SB at its middle point 0. § 123 

II. When the given point is at the end of the line. 
Let B be the given point. (Fig. 2). 

To erect a J^ to the line AB at B, 

Oonstniotion. Take any point C without AB ; and from C 
as a centre, with the distance GB as a radius, describe an arc 
intersecting AB at E. 

Draw EC, and prolong it to meet the arc again at D. 

Join BB, and BD is the ± required. 

Proof. The /LB\& inscribed in a semicircle, and is therefore 

a right angle. § 264 

Hence BD is X to AB, q. e. f. 



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PROBLEMS. 107 



Proposition XXII. Problem. 

272. From a point without a straight line, to let 
fall a perpendicular upon that line. 



•s. 



/ 



^ H * -- A . '<K- ^ 

Jjot AB be the given straight line, and C the given 
point without the line. 

To let fall a 1. to the line ABfrom the point O. 

Oonstrnotion. From (7 as a centre, with a radius sufficiently 
great, describe an arc cutting AB in two points, JJand K. 

From -ffand JTas centres, with equal radii greater than ^HK, 

describe two arcs intersecting at 0. 

Draw CO, 

and produce it to meet AB at M. 

CMia the X required. 

Proof. Since Cand are two points equidistant from JSTand 
Kf they determine a X to HK eX its middle point. § 123 

Q.E.F. 



Note. Oiven lines of the figures are full lines, rewlUng lines are 
long-dotted, and auxiliary lines are short-dotted. 



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108 PLANE GEOMETRY. — BOOK U. 

Proposition XXIII. Problem. 
273. To bisect a given straight line. 



Let AB he the given straight line. 

To bisect the line AB. 

Oonstraction. From A and B as centres, with equal radii 
greater than ^ AB, describe arcs intersecting at C and E. 

Join CE, 

Then the line C£? bisects AB. 

Proof. C and E are two points equidistant from A and B. 
Hence they determine a J. to the middle point of AB, § 123 

aE.F. 



Ex. 87. To find in a given line a point X which shall be equidis- 
tant from two given points. 

Ex. 86. To find a point X which shall be equidistant from two 
given points and at a given distance from a third given point. 

Ex. 89. To find a point X which shall be at given distances from 
two given points. 

Ex. 90. To find a point X which shall be equidistant from three 
given points. 



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PROBLEMS. 



109 



Proposition XXIV. Problem. 



274. To bisect a given arc. 




Let ACB be the given arc. 

To bisect the arc ACB, 

Oonatruotion. Draw the chord AB. 

From A and B as centres, with equal radii greater than 
\ AB^ describe arcs intersecting at D and E, 

Draw BE. 

i)-& bisects the arc ACB. 

Proof. Since B and E are two points equidistant from A 
and B, they determine a JL erected at the middle of chord 
AB. §123 

And a ± erected at the middle of a chord passes through 
the centre of the O, and bisects the arc of the chord. § 234 

Q.E.F. 

Ex. 91. To constract a circle having a given radios and passing 
through two given points. 

Ex. 92. To constmct a circle having its centre in a given line and 
passing through two given points. 



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110 PLANE GEOMETRY. — BOOK II. 

Proposition XXV. Problem. 
276. To bisect a given angle. 




Let AEB be the given angle. 

To bisect A AEB. 

OonBtruction. From jE7 as a centre, with any radius, as EA, 
describe an arc cutting the sides of the A E 2X A and B. 

From A and B as centres, with equal radii greater than 
one-half the distance from A to B, describe two arcs inter- 
secting at C. 

Join EC, AC, and BC 

EC bisects the Z E. 

Proof. In the A ^^(7 and BEC 

AE= BE, and AC= BO, Cons. 

and EC-^EC Iden. 

..AAEC^ABEC, §160 

{having three sides equal each to each). 

,'.ZAEC=ZBEC 

Q. E. F. 



Ex. 93. To divide a right angle into three equal parts. 
Ex. 94. To construct an equilateral triangle, having given one side. 
Ex. 95. To find a point X which shall be equidistant from two given 
points and also equidistant from two given intersecting lines. 



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PROBLEMS. Ill 



Proposition XXVI. Problem. 

276. At a given, point in a given straight line, to 
construct an angle equal to a given angle. 





Let C be the given point in the given line CM, and 
A the. given angle. 

To construct an /L at O eqyual to the A A. 

Oonstrnotion. From -4 as a centre, with any radius, as AE, 
describe an arc cutting the sides of the /. AdX E and F. 

From (7 as a centre, with a radius equal to ^-B, 

describe an arc cutting OMdX H, 

From H 2A2i, centre, with a radius equal to the distance EF, 

describe an arc intersecting the arc HO at m. 

Draw Om, and HCm is the required angle. 

Proof. The chords -EjPand Hm are equal. Cons. 

.-. arc EF= arc Hm, § 230 

(tn equal © equal chorda subtend equal arcs), 

.'.ZC=ZA, §229 

(tn eqy^l ® equal arcs subtend equal A at the centre). q. e. p. 



Ex. 96. In a triangle ABC, draw DE parallel to the base BC, cut- 
ting the sides of the triangle in D and E, so that DE shall equal 
DB^EC 

Ex. 97. If an interior point of a triangle ABC'vr joined to, the ver- 
tices B and (7, the angle BOC is greater than the angle BAC of the 
triangle. 



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112 PLANE GEOMETRY. — BOOK II. 



Pboposition XXVII. Problem. 

277. Two angles of a triangle bein$ given, to find 
the third angle. 




! 



E- 




U 

Let A and B be the two given angles of a triangle. 

To find the third Z of the A, 

Oonstrnction. Take any straight line, as EF, and at any 
point, as JTj 

construct Z a equal toZA^ § 276 

and Z b equal to ZB. 
Then Z c is the Z required. 

Proof. Since the sum of the three ^S of a A = 2 rt. i4, § 138 
and the sum of the three A a, i, and c, = 2rt.A; § 92 
and since two A of the A are equal to the A a and 6, 
the third Z of the A will be equal to the Z c. Ax. 3. 



aE.F. 

Ex. 98. In a triangle ABQ given angles A and B, equal respectively 
to 37° 13' 32^' and 41° 17' 66''. Find the value of angle 0. 



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PROBLEMS. 113 



Proposition XXVIII. Problem. 

278. Thratigh a given point, to draw a straight line 
parallel to a given straight line. 




D 

Let AB be the given line, and C the given point. 

To draw through the point Oa line parallel to the line AB, 
Chmstniction. Draw DCE^ making the Z. EDB, 

At the point C construct Z EOF= Z EDB. § 276 
Then the line FCHia II to AB. 

Proot Z ECF= Z EDB. Cons. 

,\HFi^\\ioAB, §108 

{when two straight lines, lying in the same plane, areciUbya third straight 
linet if the ext.-int, A are equal, the lines are parallel). 

Q. E. F. 



Ex. 99. To find a point X equidistant from two given points and 
also equidistant from two given parallel lines. 

Ex. 100. To find a point X equidistant from two given intersecting 
lines and also equidistant from two given parallels. 



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114 



PLANE GEOMETRY. — BOOK 11. 



Proposition XXIX. Problem. 

279. To divide a given straight line into equal 
parts. 

A^ ^-:. 7 tB 




Let AB he the given straight line. 

To divide AB into equal parU, 

Oonstrtiction. From A draw the line AO. 

Take any convenient length, and apply it to -40 as many 
times as the line AB is to be divided into parts. 

From the last point thus found on AO, as C, draw CS. 

Through the several points of division on -40 draw lines 
II to CB, and these lines divide AB into equal parts. 

Proof. Since -40 is divided into equal parts, AB is also, § 187 

(i^ thru or more lU intercept equal parts on any trangoenal^ they intercept 
equal parts on every transversal), 

Q.E.F. 



Ex. 101. To divide a line into four equal parte by two different 
methods. 

Ex. 102. To find a' point Xin one eide of a given triangle and equi* 
distant from the other two sides. 

Ex. 103. Through a given point to draw a line which shall mak« 
•qnal angles with the two sides of a given angle. 



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PROBLEMS. 115 



Pboposition XXX. Peoblbm. 

280. Turn sides and the included angle of a tricmr 
gle being given, to construct the triangle. 

n 



b 







./ ' 



^zl 



Let the two sides of the triangle be b and c, and the 
included angle A. 



To construct a A having two sides equal to b and c respec- 
tively, and the included Z.=^/L A. 

Chmstrnotion. Take AB equal to the side c. 

At A, the extremity of AB, construct an angle equal to the 

given Z 4. §276 

On AD take AC equal to 6. > 

Draw CB. 

Then A -4 CB is the A required. 

Q.E.F. 



Ex. 104. To constract an angle of 45^. 

Ex. 106. To find a point X which shall be equidistant from two 
given intersecting lines and at a given distance from a given point. 

Ex. 106. To draw through two sides of a triangle a line || to the 
third side so that the part intercepted between the sides shall have a 
given length. 



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116 PLANE GEOMETRY. — BOOK II. 



Proposition XXXI. Problem. 

281. A side and two angles of a triangle being 
given, to construct the triangle. 




E^ 1-9.. 

c /" 



Let c be the given side, A and B the given angles. 

To construct the triangle, 

Oonstniotion. Take J57(7 equal to e. 

At the point U construct the ZC^JST equal to Z ^. § 276 

At the point C construct the Z -BCiT equal to Z B. 

Let the sides EHB,nA. GST intersect at 0. 
Then A COE is the A required. 

Q.E.R 

Remabk. If one of the given angles is opposite to the given side, 
find the third angle by J 277, and proceed as above. 

Disonssion. The problem is impossible when the two given 
angles are together equal to or greater than two right angles. 



Ex. 107. To construct an angle of 150**. 

Ex. 108. A straight railway passes two miles from a town. A place 
is four miles from the town and one mile from the railway. To find by 
construction how many places answer this description. 

Ex. 109. If in a circle two equal chords intersect, the segments of one 
chord are equal to the segments of the other, each to each. 

Ex. 110. AB is any chord and AQ\& tangent to a circle at A, CDE a 
line cutting the circumference in D and E and parallel to AB\ show 
that the triangles ACD and EAB are mutually equiangular. 



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PROBLEMS. 117 



Peopositioh XXXII. Pboblsh. 

282. niie three sides of a triangle being given, to 
construct the triangle. 

'" .0/"" 



% 



/ X 

n/ \m 



^^ _ N.B 

Let the three sides be m, n, and o. 

To construct the triangle, 

Oonstraotion. Draw AB equal to o. 

From ^ as a centre, with a radius equal to n, describe an 
arc; 

and from J? as a centre, with a radius equal to m, describe 
an arc intersecting the former arc at 0. 

Draw CA and OB, 

Then A CAB is the A required. 

Q.E.F. 

Disonssion. The problem is impossible when one side is equal 
to or greater than the sum of the other two. 

Ex. 111. The base, the altitude, and an angle at the base, of a tri- 
angle being given, to construct the triangle. 

Ex. 112. Show that the bisectors of the angles contained by the oppo- 
site sides (produced) of an inscribed quadrilateral intersect at right angles. 

Ex. 113. Given two perpendiculars, AB and CD, intersecting in 0, and 
a straight line intersecting these perpendiculars in E and F\ to construct 
a square, one of whose angles shall coincide with one of the right angles 
at O, and the vertex of the. opposite angle of the square shall lie in £!F, 
(Two Bolutions.) 



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118 



PLANE GEOMETRY. — BOOK II. 



Pboposition XXXni. Problem. 

283. Two sides of a triangle and the angle opposite 
one of them being given, to construct the triangle. 

P 




4\ 



y 



\a 



Jl^B 



Case I. If the side opposite the given angle is less than the 
other given side. 

Let b be greater than a, and A the given angle. 

To construct the triangle. 

OonBtmotioiL Construct Z DAE = to the given Z A. ^ 276 
On AB take AB = b. 
From ^ as a centre, with a radius equal to a, 
describe an arc intersecting the line AE eii (7 and C". 
Drawee and ^C". 

Then both the A ABC and ABC^ .,i) 

fulfil the conditions, and hence we 
have two constructions. This is 
called the ambiguous case. 

Disonssion. If the side a is equal __ 
to the X BHy the arc described from 
B will touch AE, and there will be 
but one construction, the right tri- 
angle ABH. 

If the given side a is less than the 
± from B, the arc described from B 
will not intersect or touch AE, and — 
hence the problem is impossible. 



^ 



by 



X 



y 



\a 



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THE CIRCLE. 119 

If the Z ^ is right or obtuse, the problem is impossible ; for the 
side opposite a right or obtuse angle is the greatest side. § 159 

Case II. If a is equal to b. 

If the Z ^ is acute, and a = 5, the arc described from B as 
a centre, and with a radius equal to a, will 
cut the line AE bX the points A and C, 
There is therefore but one solution: the 
isosceles A ABO. 

Discussion. If the Z.A\& right or obtuse, 
the problem is impossible ; for equal sides of a A have equal 
A opposite them, and a A cannot have two right A or two 
obtuse A, 

Case III. If a is greater than b. 

If the given Z ^ is acute, the arc described from B will cut 
the line ED on opposite sides of ^, at (7 and C\ The A ABC 
answers the required conditions, but the ^/ 

A AB(y does not, for it does not contain . a.^^^^ / 

the acute Z A. There is then only one E \r'/ ^^yh^^ 

solution ; namely, the A ABC. '^*, *' 



B 



If the Z ^ is right, the arc described 
from B cuts the line ED on opposite A 

sides of A, and we have two equal right y jb^^ 

A which fulfil the required conditions. E Qf-^^,^.-'^o ^ 

If the Z.Ai8 obtuse, the arc described 

from B cuts the line ED on opposite \^ 

sides of Ay at the points C and C". The ^"^ X'^ • 

A ABC answers the required conditions, j, \/' yy* jy 

but the A ABC does not, for it does ^'"" -^'"^ 

not contain the obtuse Z A. There is then only one solu- 
tion ; namely, the A ABC. 

Q.E.F. 



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120 PLANE GEOMETRY. — BOOK II. 



Proposition XXXIV. Problem. 

284 Two sides and an included angle of a paral- 
lelogram being given, to construct the parallelogram, 

B/. :2:^ 



'A 



Let m and o be the two sides, and C the included 
angle. 

To construct a parallelogram. 

Oonstruction. Draw AB equal to o. 

At A construct the Z A equal to Z C, § 276 

and take .4^ equal tp m. 

From JJas a centre, with a radius equal to o, describe an arc. 

From 5 as a centre, with a radius equal to m, 

describe an arc, intersecting the former arc at E. 

Draw ^iTand EB, 

The quadrilateral ABEH'\% the O required. 

Proof. AB=EE, Cons. 

AH^ BE. Cons. 

.-. the figure ABEE'ib a O, § 183 

(having its opposite sidea equal). 

at F. 



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PROBLEMS. 121 



Proposition XXXV. Problem. 

286. To cirowmscribe a circle about a given tri- 
angle. ^ A 

/ 




Let ABC be the given triangle. 

To circumscribe a circle about ABC, 



Bisect AB and BO, § 273 

At the points of bisection erect Js. § 271 

Since BO is not the prolongation of AB^ these Ja will in- 
tersect at some point 0. 

From 0, with a radius equal to OB^ describe a circle. 

O ABO is the required. 

Proof. The point is equidistant from A and B, - 

and also is equidistant from B and (7, § 122 

{every point in the J. erected at the middle of a straight line is equidistant 
from the extremities of that line). 

.*. the point is equidistant from Aj B, a,nd O, 

and a O described from as a centre, with a radius equal to 
OB, will pass through the vertices A, B, and 0. atF. 

286. Scholium. The same construction serves to describe a 
circumference which shall pass through the three points not 
in the same straight line ; also to find the centre of a given 
circle or of a given arc, 



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V 



122 PLANE GEOMETRY. — BOOK II. 



Proposition XXXVI. Problem. 

287. Through a given point, to draw a tangent to a 
given circle. 





^--A---' 



Case I. When the given point is on the circle. 
Let C be the given point on the circle. 

To draw a tangent to the circle at O. 

OoiiBtruotion. From the centre draw the radius 00. 

- Through O draw AM± to 00 § 271 

Then A Mis the tangent required. 
Proof. A straight line ± to a radius at its extremity is tan- 
gent to the circle. § 239 

Case II. When the given poirvt is without the circle. 
Let be the centre of the given circle, E the given 
point without the circle. 

To draw a tangent to the given circle from the point JS. 
Oonstmction. Join OE. 

On OU as a diameter, describe a circumference intersecting 
the given circumference at the points -3f and -H". 
Draw Oif and EM. 
Then EM is the tangent required. 

Proof. Z OME is a right angle, § 264 

Qteing inscribed in a semicircle), 

.'. EMia tangent to the circle at M, § 239 

In like manner, we may prove SE tangent to the given O. 

Q.E.F. 



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PROBLEMS. 123 

Proposition XXXVII. Problem. 
28& To inscribe a circle in a given triangle. 




Let ABC be the given triangle. 
To inscribe a circle in the A ABC, 

Oonstmotion. Bisect A A and (7. § 275 

From E^ the intersection of these bisectors, 

draw EHJL to the line AC. § 272 

From E, with radius EH, describe the O KMH, 
The O KB:M\& the O required. 
Proof • Since E is in the bisector of the Z ^, it is equidis- 
tant from the sides AB and AC\ and since E is in the bisector 
of the Z, C, it is equidistant from. the sides ^(7 and BC, § 162 
{every point in the bisector of an Z is equidistant from the sides of the Z). 
.'. a O described from E as centre, with a radius equal to 
EH, will touch the sides of the A and be inscribed in it. 



aE.F. 



289. Scholium. The intersec- 
tions of the bisectors of exterior 
angles of a triangle, formed by 
producing the sides of the tri- 
angle, are the centres of three 
circles, each of which will touch . 
one side of the triangle, and the 
two other sides produced. These 
three circles are called esaibed 
circles. 



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124 PLANE GEOMETRY. — BOOK IL 



Peoposition XXXVIII. Problem. 

290. Upon a given straight line, to describe a seg" 
ment of a circle which shall contain a given angle. 



Let AB be the given line, and M the given angle. 
To describe a segment upon AB which shall contain A M, 

Oonstraotion. Construct Z^^^ equal to Zitf! §276 

Bisect the line AB by the ± FO. § 273 

From the point B draw BO ± to EB, § 271 

From 0, the point of intersection of FO and BO, as a cen- 
tre, with a radius equal to OB, describe a circumference. 
The segment AKB is the segment required. 

Proof. The point is equidistant from A and J9, § 122 
(every ipoini in a 1. erected at the middle of a straight line is equidistant 
from the extremities of that line). 

.*. the circumference will pass through A, 
But BF is ± to OB. Cons. 

/. BF is tangent to the O, § 239 

• (a straight line ±to a radius at its extremity is tangent to the 0). 

.-. Z ABF is measured by i arc AB, § 269 

(being an Z formed by a tangent and a chord). 

An Z inscribed in the segment AKB is measured by 

\AB. §263 

,'. segment AKB contains A ]i£. Ax. 1 



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PROBLEMS. 125 



Proposition XXXIX. Problem. 

291. TofiTi^theraMooftwo(^Tnw£'nsur(M 
lines. 

J. 1 i^ 

(^ 1 1 r^B 

F 

Let AB and CD be two straight lines. 

To find the ratio of AB and CD. 

Apply CDM AB as many time^as possible. 

Suppose twice, with a remainder EB, 

Then apply EB to CD as many times as possible. 

Suppose three times, with a remainder FD. 

Then apply FD to EB as many times as possible. 

Suppose once, with a remainder HB, 

Then apply J7B to FD as many times as possible. 

Suppose once, with a remainder KD. 

Then apply KD to HB as many times as possible. 

Suppose KD is contained just twice in HB, 

The measure of each line, referred to KD as a unit, will 
then be as follows : 

HB = 2KD) 

FD= HB-\-KD= ZKD) 
EB= FD+BrB= 5KD; 
CD =SEB + FD = 18KD; 
AB=2CD +EB^^1KD; 
• AB^ ^IKD , 
. "CD IS KD' 
AB_ 
CD 18 aE.F. 



.the ratio 4£ = ~ 



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126 PLANE GEOMETRY. — BOOK II. 



Theorems. 

114. The shortest line and the longest line which can be drawn from 
a given exterior point to a given circumference pass through the centre. 

116. If through a point within a circle a diameter and a chord X to 
the diameter are drawn, the chord is the shortest cord that can be drawn 
through the given point. 

116. In the same circle, or in equal circles, if two arcs are each 
greater than a semi-circamference, the greater arc subtends the ksB 
chord, and conversely. 

117. If ABC is an inscribed equilateral triangle, and Pis any point 
in the arc J?C, then PA = FB + PC. 

Hint. On FA take Pif equal to PP, and join BM. 

^ 118. In what kinds of parallelograms can a circle be inscribed? 
Prove your answer. 

> 119. The radius of the circle inscribed in an equilateral triangle is 
equal to one- third of the altitude of the triangle. 

120. A circle can be circumscribed about a rectangle. 

121. A circle can be circumscribed about an isosceles trapezoid. 

jt^ 122. The tangents drawn through tiie vertices of an inscribed rec- 
tangle enclose a rhombus. 

-f-^ 123. The diameter of the circle inscribed in a rt. A is equal to the 
difference between the sum of the legs and the hypotenuse. 

124. From a point A without a circle, a straight line AOB is drawn 
through the centre, ?ind also a secant ACD, so that the part AC without 
the circle is equal to the radius. Prove that Z DAB equals one-third 
the Z DOB. 

125. AH chords of a circle which touch an interior concentric circle 
Sbfe equal, and are bisected at the points of contact. \ 

'^ 126. If two circles intersect, and a secant is drawn throu^ each 
point of intersection, the chords which join the extremities of the sd<jant8 
are parallel. Hint. By drawing the common chord, two inscrft^ed 
quadrilaterals are obtained. 

127. If an equilateral triangle is inscribed in a circle, the distance of 
each side from the centre of the circle is equal to half the radius. 

128. Through one of the points of intersection of two circles a 
diameter of each circle is drawn. Prove that the straight line joining 
the ends of the diameters passes through the other point of intersection. 



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EXERCISES. 127 

129. A circle toaches two sides of an angle £AC&t B, C\ through any 
point D in Uie arc BC a tangent is drawn, meeting AB at E and AQ 
at F. Prove (i.) that the perimeter of the triangle -4^^ is constant for 
all positions of i> in BC\ (ii.) that the angle EOF'vi also constant 

Loci. 

130. Find the locns of a point at three inches from a given point. 

131. Find the locns of a point at a given distance from a given 
circumference. 

132. Prove that the locns of the vertex of a right triangle, having a 
given hypotennse as hase, is the circumference described upon the given 
hypotenuse as diameter. 

Prove that the locus of the vertex of a triangle, having a given 
/Chase and a given angle at the vertex, is the arc which forms with the 
base a segment capable of containing the given angle. 

134. Find the locus of the middle points of all chords of a given 
length that can be drawn in a given circle. 

135. Find the locus of the middle points of all chords that can be 
drawn through a given point J. in a given circumference. 

iJ 136. Find the locus of the middle points of all straight lines that can 
|e drawn from a given exterior point -4 to a given circumference. 

137. A straight line moves so that it remains parallel to a given line, 

and touches at one end a given circumference. Find the locus of the 

^^er end. 

\r7ll38. A straight rod moves so that its ends constantly touch two 

wed rods which are X to each other. Find the locus of its middle point. 

139. In a given circle let AOB be a diameter, OC any radius, CD 
the perpendicular from Cto AB. Upon OCtake OU'^QD. Find the 
locus of the point if as 0(7 turns about 0. 

Construction op Polygons. 

To construct an equilateral A, having given : 
7 140. The perimeter. ^141. The radius of the circumscribed circle. 
)142. The altitude. ,143. The radius of the inscribed circle. 

To construct an isosceles triangle, having given; 
144. The angle at the vertex ftad ihe basa 



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128 PLANE GEOMETRY. — BOOK II. 

145. The angle at the vertex and the altitnde. 

146. The base and the radius of the circumscribed circle. 

147. The base and the radios of the inscribed circle. 

148. The perimeter and the alti- 
tnde. 

Hints. Let ABO be the A re- C 

quired, and EF the given perimeter. 
The altitude CD passes through the y<' 

middle of EF, and the &. AEC, 
BFGwe isosceles. e" 

To construct a right triangle, having given : 

149. The hypotenuse and one leg. 

150. The hypotenuse and the altitude upon the hypotenuse. 

151. One leg and the altitude upon the hypotenuse as base. 

^SJ.52. The median and the altitude drawn from the vertex of the rtZ. --C^ 

153. The radius of the inscribed circle and one leg. 

1^4. The radius of the inscribed circle and an acute angle. 

y 155. An acute angle and the sum of the legs. 

'156. An acute angle and the difference of the legs. . -^ 

To construct a triangle, having given : 

V157. The base, the altitude, and the Z at the vertex. 

/^158. The base, the corresponding median, and the Z at the vertex. 

159. The perimeter and the angles. « :^ , 

> 160. One side, an adjacent Z, and the sum of the other sides. 

161. One side, an adjacent Z, and the difference of the other sidesA 

^ 162. The sum of two sides and the angles. 

7163. One side, an adjacent Z, and radius of circumscribed O. 

-164. The angles and the radius of the circumscribed 0. 

' 165. The angles and the radius of the inscribed O. 

'166. An angle, the bisector, and the altitude drawn from the vertex. 

167. Two sides and the median corresponding to the other side. 

168. The three medians. 

To construct a square, having given : 

169. The diagonal. 17^ '^I'he sum of the diagonal and one side. 



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EXERCISES. 129 

To constract a rectangle, having given : 
]A*IL One side and the Z formed by the diagonals. 
**r 172. The perimeter and the diagonal. 
*^^- 173. The perimeter and the Z 'of the diagonals. 

174. The difference of the two adjacent sides and the Z of the 
diagonals. 

To constmct a rhombus, having given: 

>.175. The two diagonals. 

yl76. One side and the radius of the inscribed circle. 
^ 177. One angle and the radius of the inscribed circle. 
^^>178. One angle and one of the diagonals. 

To construct a rhomboid, having given: 
^-179. One side and the two diagonals. 
' 180. The diagonals and the Z formed by them. 
181. One side, one Z, and one diagonal. 
^182. The base, the altitude, and one angle. 

To constract an isosceles trapezoid, having given: 
"^183. The bases and one angle. 184. The bases and the altitude. 

185. The bases and the diagonal. 

186. The bases and th. radius of the circumscribed circle. 

To constract a trapezoid, having given : 
"^ 187. The four sides. 188. The two bases and the two diagonals. 
j^^ The bases, one diagonal, and the A formed by the diagonals. 

Construction op Circles. 

l^d the loou9 of the centre of a circle : 

190. Which has a given radius r and passes through a given point P. 

191. Which has a given radius r and touches a given straight line AB» 

192. Which passes through two given points Pand Q. 

193. Which touches a given straight line AB at a given point P. 

194. Which touches each of two given parallels. 

196u Which touches each of two given intersecting liuea^ < 



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130 PLANE GEOMETRY. — BOOK H. 

To construct a circle which has the radios r and which alao : 

196. Touches each of two intersecting lines AB and CD, 

197. Touches a given line AB and a given circle K, 

198. Passes through a given point *P and touches a given line AB. 

199. Passes through a given point P and touches a given circle K. 

To construct a circle which shall : 

200. Touch two given parallels and pass through a given point P. 

201. Touch three given lines two of which are parallel 

202. Touch a given line AB at P and pass through ^ given point Q. 

203. Touch a given circle at P and pass through a given point Q. 

204. Touch two given lines and touch one of them at a given point P. 
> 205. Touch a given line and touch a given circle at a point P. 
v206. Touch a given line AB at P and also touch a given circle. * 

-^7. To inscribe a circle in a given sector. 

^208. To construct within a given circle three equal circles, so that 
each shall touch the other two and also the given circle. 
^>209. To describe circles about the vertices of a given triangle as 
centroB, so that each shall touch the two others. 



Construction of Straiqat Lines. 

y 210. To draw a common tangent to two given circles. 

211. To bisect the angle formed by two lines, without producing the 
lines to their point of intersection. 

2 212. To draw a line through a given point, so that it shall form with 
/Che sides of a given angle an isosceles triangle. 
,J^ 213. Given a point P between the sides of an angle BAC. To draw 
through P a line terminated by the sides of the angle and bisected at P. 

Z214. Given two points P, Q, and a line AB ; to draw lines from P 
i Q which shall meet on AB and make equal angles with AB, 
Hint. Make use of the point which forms with P a pair of points 
symmetrical with respect to AB. 

7 216. To find the shortest path from Pto Q which shall touch a line AB, 
^ 216. To draw a tangent to a given circle, so that it shall be parallel 
U a given straight line. 



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BOOK ni. 

PROPORTIONAIi LINES AND SIMILAR 
POLYGONS. 



The Theory of Pbopobtion. 

292. A proportion is an expression of equality between two 
equal ratios. 

A proportion may be expressed in any one of the follow- 
ing forms : 

7=^; a:b = c:d; a:b::c:d; 
a 

and is read, " the ratio of a to 6 equals the ratio of o to ^.** 

293. The terma of a proportion are the four quantities com- 
pared ; the first and third terms are called the antecedents, the 
second dLiii fourth terms, the conseqtcents ; the first and fourth 
terms are called the extremes, the second and third terms, the 
means, 

294. In the proportion a:b = c:dy d is a fourth proper' 
tional to a, b, and c. 

In the proportion a : J = 6 : c, c is a third proportional to 
a and b. 

In the proportion a:b=^b:c, b is a m^an proportional 
between a and c. 



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132 PLANE GEOMETRY. — BOOK ni. 



Peoposition I. 

295. In every proportion the product of the extremes 
is equal to the product of the means. 

Let aib=:cid. 

To prove cui = bc. 

Now f=2, 

a 

whence, by multiplymg both sides by bd, 

ad=^be. g.ta 

Proposition II, 

296. ^ mean proportional betiveen ttvo puantiiiss 
is e^qual to the square root of their product. 

In the proportion a:b = b\c, 

V^ac, §295 

(tJie product of the extremes is eqiml to the product of the means). 

Whence, extracting the square root, 

Proposition III. 

297. If the prodvAit of two quantities is equal to the 
product of two others, either two may be made the 
extremes of a proportion in which the other two are 

made the means. 

Let ad = be. 

To prove a:b = c:d. 

Divide both members of the given equation by bd. 

Then ?--=% 

b d 



or, a:b = c:d. 



aE.a 



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THEORY OF PBOPOBTION, 133 



Proposition IV. 

298. If four quantities of the same kind are in pro- 
portion, they will he in proportion by alternation; 
that is, the first term JvUl be to the third as the seo- 
ond to the fourth. 

Let aib^cid. 

To prove a:c = b:d. 

Now 2 = ^. 

b a 

Multiply each member of the equation by -. 

Then ?=t 

c d 



or, a\c = b:d. 



aft.a 



Proposition V. 

299. If four quantities are in proportion, they will 
be in proportion by inversion ; that is, the second term 
will be to the first as the fourth to the third. 

Let aih^cid* 

To prove b:a = d:e. 

Now bc = ad. § 295 

Divide each member of the equation by ao. 

Then i=^, 

a c 



pr, b:a = d:o. 



at.a 



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134 PLANE QEOMBTRY. — BOOK III. 

Proposition VI. 

8OO1 If four quantities are in proportion, they vrill 
be in proportion by composition ; tJuit is, the sum of 
the first two terms wUl be to the second term as the 
sum, of the last two terms to the fourth term. 
Let a:b = c:d. 

To prove a + b:b = c + d:(L 

-rs 

Add 1 to each member of the equation. 



Then 


f+'-j+i^ 


that is, 


a+b c+d 
h d • 


or, 


a + b:b = c + d:d. 


In like manner, 


a + b:a=c + d:c. 




Proposition VII. 



Q.E.a 



SOL If four quantities are in proportion, they will 

be in proportion by division ; tha^ is, the difference 

of the first two terms wUl be to the second term as 

the difference of the last two terms to the fourth 

term. 

Let a : 5 = e : d. 

Topi'ove a — b:b = c — d:d. 

Subtract 1 from each member of the equation. 

Then «- 1=^-1; 

b d 



a — b __ c — d 
b d ' 

or, a — b:b = c — d:d. 



that is, 

In like manner, a-'b:a=e — d:o. ctta 

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THEORY OP PROPORTION. 135 

Proposition VIII. 

802. In any proportion the terms are in proportion 
by composition and division; that is, the sum of the 
first two terms is to their difference as the sam of 
the last two terms to their difference. 
Let a:b = c:d, 

a + b _ c-\-d 

a c 

a — h c — ct 



Then, by § 300, 
And, by § 301, 
By division, 



a c 

a + b _ c + d 
a—b c—d 



or, a+ft:a — 6=<? + rf:e? — €?• 



Proposition IX. 



acD. 



303. In a series of equal ratios, the sum of the an- 
tecedents is to the sum of the consequents as any 
antecedent is to its consequent. 

Let a: 6 = c:d=c:/=or:^ 

Toprove a+c+e+g:b + d+f+h = a:b. 

Denote each ratio by r. 

Whence, a=^br^ c^dr^ «=y^, ^ = Ar. 
Add these equations. 

Then a + (? + <5 + ^ = (i + <Z+/+ A)r. 

Divide by (6 + c?+/+ A). 

Then « + £±l±2=^ = «, 

*+rf+/+A b 

or, a+(?+c+a:i + rf+/+A = a:i. 

as-D. 



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136 PLANE QEOMETRY. — BOOK IH. 

Proposition X. 

804 The products of the corresponding terms of 
two or more proportions are in proportion. 
Let a:b = e:d, e:f=gih, kil^nnitu 

To prove aek:bfl = cgm:dhfL 

Now 5 = ^, ^=2, ^=??. 

b d f h I n 

Whence, by multiplication, 

aek^cgm 

bfl dhn 

or, aek:bfls=cgm:dhiu 

Proposition XI. 



ctca 



806. Like powers, or like roots, of the terms of a 
proportion are in proportion. 

Let aih=.cid. 
To prove a^:b^=^(f^:d\ 

ill* 
and Of* :b» = c» : c?». 

Now f=^ 

b d 

By raising to the nth power, 

^ = ^; OT ar:b^ = (r:d* 
0* d^ 

By extracting the nth root, 

J (^ 1 ,1 1 ,1 

— = — ; or, a» : 6» = c* : a». 

b^ rfn 

Q.E.a 
806t Equimultiples of two quantities are the products ob- 
tained by multiplying each of them by the same number. 
Thus, ma and mi are equimultiples of a and b. 



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THEORY OP PBOPORTION. 137 



Proposition XII. 

307. Equimultiples of two quantities are in the 
same ratio as the quantities themselves. 

Let a and b be any two quantities. 

To prove ma : mb =a:b. 

Now f=f. 

Multiply both terms of first fraction by m. 

Then ^=?, 

mo b 

or, m>a : mb = a : 6, 

ae.D. 



308i Scholium. In the treatment of proportion it is as- 
sumed that fractions may be found which will represent the 
ratios. It is evident that the ratio of two quantities may be 
represented by a fraction when the two quantities compared 
can be expressed in irUegers in terms of a common unit. But 
when there is no unit in terms of which both quantities can be 
expressed in integers^ it is possible to find a fraction that will 
represent the ratio to any required degree of accuracy, (See 
§§251-256.) 

Hence, in speating of the product of two quantities, as for 
instance, the product of two lines, we mean simply the product 
of the numbers which represent them when referred to a com- 
m>on unit. 

An interpretation of this kind must be given to the product 
of any two quantities throughout the Geometry. 



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138 PLANE GEOMETRY. — BOOK in. 

Proportional Lines. 

Proposition I. Theorem. 

809. If a line is drawn through two sides of a tri' 
angle parallel to the third side, it divides those sides 
proporUonally. 

A A 




EjL \F .^/- -^ 




M/. _ \ KV 

p/ " " "^ 

B Fig. 1. C Fig. 2. 

In the triangle ABC let EF "be dz^wn parallel to BC. 

To prove ^=^' 

^ AE AF 

Case I. When AE and EB (Fig. 1) are commensurable. 

Find a common measure of AE and EB, as BM, 

Suppose BM to be contained in BE three times, 

and in AE four times. 

Then ^=§. m 

AE 4 ^ ^ 

At the several points of division on BE and AE draw 
straight lines li to ^C7. 

These lines will divide AC into seven equal parts, of which 
i^(7 will contain three, and -4/ *vill contain four, §187 

(if pa^raUeh intercept equal parts on any tra^isversalj they intercept equal 
parts on every transversal), 

.•.:^ = ?. (2) 

AF 4 ^ ^ 



Compare (1) and (2), 

EB 



AE A. ^^^- 



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PROPORTIONAL LINES. 139 

Case II. When AE and EB (Fig. 2) are mcommenmrahle. 

Divide AE into any number of equal parts, and apply one 
of these parts as a unit of measure to EB as many times as it 
will be contained in EB, 

Since AE and EB are incommensurable, a certain number 
of these parts will extend from ^ to a point JST, leaving a 
remainder KB less than the unit of measure. 
Draw^^lltoJ^C. 

Then EK^FH^ ^ j 

AE AF 

Suppose the unit of measure indefinitely diminished, the 

ratios -— - and -— =: continue equal ; and approach indefi- 
AE AF ^ ^^ 

EB F*C! 

nitely the limiting ratios -— and -— ;, respectively. 

Therefore ^=1^. §260 

AE AF 

aE.0. 

310. CoR. 1. One aide of a triangle is to either part cut off 
hy a straight line parallel to the base as the other side is to the 
corresponding part 

For EB : AE= FC: AF, by the theorem. 

,\ EB + AE: AE= FC+ AF: AF, §300 

or AB:AE=AO:AF 

311. Cor. 2. ^ two lines are cxd hy any numher of parallels, 
the co7'responding intercepts are proportional. 

Let the lines be AB and CD. 

Draw -4iV II to CD, cutting the lis at L, M, 
and iV. Then 

AL=CO, LM=OK, MN=KD, §187 

By the theorem, B n D 

AE-: AM= AF: AL = FE:: LM= HB : MN. 
That is, AF:CQ= FH: QK= HB:KD. 

If the two lines AB and CD were parallel, the correspond- 
ing intercepts would be equal, and the above proportion be true. 




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140 PLANE GEOMETRY. — BOOK III. 

Pboposition II. Theorem. 

812. If a straight line divides two sides of a trir 
(ingle proportionally, it is parallel to the third side. 



In the triangle ABC let EF be drawn so that 

AB^AC 
AE AF 

To prove EF^toBO. 

Proof. From E draw EH II to BO. 

Then AB : AE= AC: AH, § 810 

{one side of a Auto either part cut off by a line II to the base, as the other 
side M to the corresponding part). 

But AB'.AE=AC:AF. Hyp. 

The last two proportions have the first three terms equal, 
each to each ; therefore the fourth terms are equal ; that is,^ 

AF= AH. 

/. -fi'JPand jBZfl" coincide. 

But -E^^is II to BO. Cons. 

.-. EFy which coincides with EH, is | to BO, 



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PROPORTIONAL LINES. 141 



Pboposition III. Theoeem. 

813. The bisector of an angle of a triangle divides 
the opposite side into segments proportional to the 
other two sides. 



\ 



-.P 




AM B 

Let CM bisect the angle C of the triangle GAB. 
To prove MA:MB=^CA: CB, 

Proof. Draw AE I) to MC to meet -BC produced at E, 

Since MO \&\\ to AE of the A BAE, we have § 309 

MA:MB^CE'.CB. (1) 

Since MC is II to AE, 

ZACM=ZCAE, §104 

(fimng oltAnt A of II lines) ; 

and Z BCM= Z OEA, § 106 

(being extAnt AoJW lines). 

But the Z A CM= Z BCM Hyp. 

.\ the ZCAE = ZCEA, Ax. 1 

/.CE=CA, §156 

(t^ two Aofat^ are equal, the opposite sides are equaX), 

Putting CA for CE in (1), we have ■ 

MA:MB=CA:CB. \ 

aE.ix 



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142 PLANE GEOMETRY. — BOOK III. 



Peoposition IV. Theorem. 

814. The bisector of an exterior angle of a triangls 
meets the opposite side produced at a point the dis- 
tances of which from the extremities of this side are 
proportional to the other two sides. 

^\ 



A 

Let CM* bisect the exterior angle ACE at the tri- 
angle CAB, and meet BA produced at MK 

To prove IP A : M'B = CA : CB, 

Proof. Draw AF II to CM' to meet BC 2X F. 
Since AFis II to CM* of the A BCM\ we have § 309 

M'A:M'B=CF:CB, (1) 

Since AFis II to CM\ 

the Z M'CE = Z AFC, § 106 

ipeing ext.-int. AofW lines) ; 

and t\ieZM'CA = ZCAF, §104 

{being altAnU A of II hnes). 

Since CM* bisects the Z EGA, 

ZM'CE=ZM*CA, 

.'. the Z AFC = Z CJ.F Ax. 1 

.-. CA = CF, § 156 

(if two A of a A are equal, the opposite sides are eqital). 

Putting CA for CF in (1), we have 

M*A:M'B=CA:CB. 

aE.a 



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PROPORTIONAL LINES. 143 

816. Scholium. If a given line AB is divided at M, a 
point between the extremities A and B, it is said to be 
divided internally into the segments MA and MB ; and if it 
is divided at M\ a point in the prolongation of AB, it is said 
to be divided externally into the segments M^A and M^B, 

"- -- -X— I ' 

In either case the segments are the distances from the point 
of division to the extremities of the line. If the line is divided 
internally, the sum of the segments is equal to the line ; and 
if the line is divided externally, the difference of the segments 
is equal to the line. 

Suppose it is required to divide the given line AB inter- 
nally and externally in the same ratio; as, for example, the 
ratio of the two numbers 3 and 5. 



M' A M B y 

We divide AB into 5 + 3, or 8, equal parts, and take 8 
parts from A ; we then have the point Mj such that 

MA:MB = S:b, (1) 

Secondly, we divide AB into two equal parts, and lay off 
on the prolongation of AB, to the left of A, three of these 
equal parts ; we then have the point M\ such that 

M'A:M'B = S:5, (2) 

Comparing (1) and (2), 

MAiMB^M'A'.WB. 

816. If a given straight line is divided internally and 
externally into segments having the same ratio, the line is 
said to be divided harmonically. 



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144 



PLA.NB GEOMETRY. — BOOK IIL 




817. Cob. 1. The bisectors of an interior angle and an exte- 
rio7' angle at one vertex of a triangle jy 
divide the opposite side harmoni- 
caUy, For, by §§ 313 and 314, each ^"^^ 

bisector divides the opposite side 
into segments proportional to the 
o^.her two sides of the triangle. 

318. Cor. 2. If the points M and M^ divide the line AB 
harmonically f the points A and B divide the line MM^ har- 
monically. 

For, if MA:MB ^ JUT A : M'B, 

by alternation, MA:M*A = MB : M'B. §298 

That is, the ratio of the distances of A from Jtf"and M^ is 
equal to the ratio of the distances of B from M and M'. 

The four points -4, B, M, and M' are called harmonic 
points, and the two pairs, A, B, and M, JIf, are called con- 
jiLgate harmonic points. 

Similar Polygons. 

319. Similar polygons are polygons that have their homol- 
ogous angles equal, and their homologous sides proportional. 




and 



ED E' 

Thus, if the polygons ABODE and A^B^C^UE^ are similar 
the AA^ -B, C7, etc., are equal to AA\ B\ G\ etc. 
AB BC CD 



A'B' BC CD 



etc. 



820. In two similar polygons, the ratio of any two homol^ 
ogous sides is called the ratio of similitude of the polygons. 



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similar tbiakgles. 145 

Similar Triangles. 
Proposition V. Theorem. 

321. Two mutually equiangular triangles are sim- 
ilar. 

A 
/\ A' 




In the triangles^ ABO and A^WC^ let angles A, B, O be 
equal to angles A\ B\ O respectively. 

To prove A ABC and A^B'C similar. 

Proof. Apply the A A'B'C to the A ABO, 

80 that Z A' shall coincide with Z A, 

Then the A A'B^O' will take the position of A AEK 

Now Z ^^^(same as ZB') = Z B. 

:. EH'i^^io BC, §108 

{when two straight lines, lying in the same plane, are cut by a third straight 
line, if the ext.-int. 4 are equal the lines are parallel). 

.-. AB : AE= AC: AH, § 310 

or AB'.A'B' = AC:A*C^, 

In like manner, by applying A A^B^C* to A ABC, so that 
Z B^ shall coincide with Z B, we may prove that 
AB.A'B^^BO'.B'O^. 

Therefore the two A are similar. § 319 

aE.D. 

322. Cor. 1. Two triangles are similar if two angles of the 
one are equal respectively to two angles of the other. 

323. Cor. 2. Two right triangles are svrmlar if an actUe 
angle of the one is equal to an acute angle of the other. 



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146 PLANE GEOMETRY. — BOOK III. 



Peoposition VI. Theorem. 

8S4. If two triangles have their sides respectively 
proportional, they are similar. 




In the triangles ABC and A'&a let 
AB ^ AC ^ BC 
A'B* A'C B^a 

To prove A ABO and A^B^C* similar. 

Proof. Take AU^ A'B\ and AH=^ A'O*. 

Draw EH. 

Then from the given proportion, 

AB^AC 
AE AH 

,'. EH is II to BO, § 312 

(i^a line divide two sides of a A proportionally, ii m II to the third side). 

Hence in the A ABOsLnd AEH 

ZABO=ZAEH, §106 

and ZAOB = ZAHE, 

(being ext.-int. AofW lines). 

.-. A ABOmd AEH are similar, § 822 

(two ▲ are similar if two A of one are equal respectively to two A of the 

other). 

.\AB: AE = BO:EH; 

that is, AB : A'B' = BO: EH. 



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SIMILAB TBIANGLES. 



147 



But by hypothesis, 

The last two proportions have the first three terms equal, 
each to each ; therefore the fourth terms are equal ; that is, 

Hence in the A ^^^and A^BC\ 

EH=B'C\ AU=A'B', s^niAS^ A'O^. 

.-. A AI!B'= A A'B'C, , § 160 

(having three tides of the one equal respectively to three iidee of the other). 

But A ABB is similar to A ABC. 

.-.A A'B'C is similar to A ABO, q. t a 

826. Scholium. The primary idea of similarity is likeness 
of form; and the two conditions necessary to similarity are : 

I. For every angle in one of the figures there must be an 
equal angle in the other, and 

II. The homologous sides must be in proportion. 

In the case of triangles, either condition involves the other, 
but in the case of other polygons, it does not follow that if one 
condition exist the other does also. 




Q' 



R' 



Thus in the quadrilaterals Q and Q!, the homologous sides 
are proportional, but the homologous angles are not equal. 

In the quadrilaterals JR and JR' the homologous angles are 
equal, but the sides are not proportional. 



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148 PLANS OSOMSTBT. — BOOK III. 



Pbopositioh VII. Theobeh. 

826. If two triangles have an angle of the one equal 
to an angle of the other, and the including sides pro- 
portional, they are similar. 




In the triangles ABC and A'B'C*, let AA^AA', and 

AB AC 
AfBf^A'C'^ 

To prove A ABC and A^B'C* similar. 

Proof. Apply the A A'B'C^ to the A ABC, so that Z A' 
shall coincide with Z A. 

Then the A A'B'C will take the position of A A£JI. 



AB _ AC 
A'B' A'C 



Now W. = -^/ Hyp. 



That is, AR = A^. -^ 

^ AE AH 

Therefore the line EH divides the sides AB and -4 ff pro- 
portionally ; 

.-. EHi^ II to BC, § 312 

(t^.a lint divide two tides of a A proportionally, it is Clothe third tide). 

Hence the A ABC and AEH are mutually equiangular 
and similar. 

.-. A A^B'C^ is similar to A ABa 

aE.a 



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SIMILAR TRIANGLES, 



149 



Proposition VIII. Theorem. 

827. If two triangles have their sides respectively 
parallel, or respectively perpendicular, they are sim^ 
ilar, 

A 

a! 





In the triangles A'BfC* and ABC let A'B', A'a, B^a he 
respectively parallel, or respectively perpendicular, 
to AB, AC, BC. ^ 

To prove . A A'£'(y and ABO siimlar. 

Proof. The corresponding A are either equal or supplements 
of each other, §§ 11^, 113 

(if two A have their sides II, or ±, they are equal or supplemen^ry). 
Hence we may make three suppositions : ^ 

1st. ^ + ^' = 2rt.^, £ + B'==2rt.A, a+C' = 2rt.A 
2d. A = A\ ^ + ^' = 2rt.A e+C' = 2rt.A 

8d. ^ = ^', ^ = ^, .-. C=(7'. §140 

Since the sum of the A of the two A cannot exceed four 
right angles, the third supposition only is admissible. § 138 

.-. theW A ABC&ni A'B*C' are similar, § 821 
{two muttLoUy equiangular ^ are nrnUar). 

Q.E.IX 



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150 



PLANE GEOMETRY. — BOOK HI. 



Proposition IX. Theoeem. 

328. The homologous altitudes of two similar tri- 
angles have the sam^e ra/tio as any two homologous 
sides. 




A o 




A' O 



In the two similar triangles ABC and A*B*a, let the 
altitudes be CO and C'C. 



To prove 



CO AC AB 



C'ff A'C A'ff 

"ttwA. In the rt. A CO A and C'aA\ 

/LA^/L A\ § 319 

{being homologous A of the similar ^ ABC and A'B'CT). 

.-. A COA and CCA' are similar, § 823 

(two H. ^ having an acute ^ of the one equal to an acute /.of the other 
are similar). 

. CO _ AO 

' ' ca A'o'' 

In the similar A ^J5Cand A'£>0', 

AC _ AB 
A'C A'Bf 

AC AB 



§819 



Therefore, 



CO 



A'C A'B' 



»*.» 



■stfc 



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SIMILAR TRIANGLES. 161 



Proposition X. Theorem. 

> 

829. Straight lines drawn through the sams point 
intercept proportional segments upon two parallels. 




Let the two parallels AE and A*E' out the straight 
lines OA, OB, OC, OD, and OE. 

^ A'B^ BO' CD' UB 

Proof. Since A'E' is II to AE, the pairs of A OAB and 
OA'B\ OBC and OB'C\ etc., are mutually equiangular and 
similar, 

, AB OB ^. BO OB ^oiQ 



(homologous sides of similar ^ are proportional). 

. AB ^ BO 
' ' A'B' BO' 

In a similar way it may be shown that 



Ax. 1 



aE.0. 



BO__^OD_ .OD__DE 
BO' O'B ^" O'B BE' 

Rehabe. a condensed form of writing the above is 

AB ( 0B\ EC ^( 0C\ CD ^( 0D\^ DE 
Afff ^ \0B')^ B'C [oaj ^ CI/ \0D') BEf' 

where a parenthesis about a ratio signifies that this ratio is used to 
prove the equality of the ratios immediately preceding and following it 



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162 PLANE GEOMETRY. — BOOK IIL 



Pboposition XI. Theorem. 

380. Conversely: If three or mare nonrparalleL 
straight lines intercept proportional segments upon 
two parotids, they pass through a oommun point. 




Let AB, CD, EF, cut the parallels AE and BP so that 

AC:BD=CE I DF. 

To prove thcU AB, CD, EF prolonged meet in a point. 

Proof. Prolong AB and CD until they meet in 0. 

Join OK 

If we designate by F the point where OE cuts BF, we 
shall have by § 329, 

AC'.BD^CEiDF'. 
But by hypothesis 

AC:BD=CE:DF 

These proportions have the first three terms equal, each to 
each ; therefore the fourth terms are equal ; that is, 

Dr = DF 

.'. F coincides with F. 

.'. J57i^ prolonged passes through O. 

/. AB, CD, and -EF prolonged meet in the point O. 

Q.E.D' 



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SIMILAR POLYGONS. 



153 



Similar Polygons. 
Proposition XII. Theorem. 

331. If two polygons are composed of the same nuwr^ 
ber of triangles, similar each to ea^h, and similarly 
placed, the polygons are similar, 
E 





B C B' C/ \^ 

In the two polygons ABODE and A'B'OD'E^y let the 
triangles AEB, BEC, CED be similar respectively to 
the triangles A^E^B^ BfE'C\ OE'DK 

To pr<yve ABODE simila/r to A^B^O^D^E. 



Proof. 

Also, 
and 

By adding, 



Z.A = ZA\ 

(being homologous A of similar A). 

ZABE=ZA'B'E\ 
ZEBO=ZEB'0'. 



319 



319 



Z ABO = Z A'B'OK 
In like manner we may prove Z BOD = Z B^O^D\ etc. 
Hence the two polygons are mutually equiangular. 
Now 

AE _ AB _( EB\_ BC___( EC\ OD _ ED 

\E'0')' 



AE _ AB __ ( EB\ 
A'E' A'B^ [e'B'J 



B^C \E'Oy O'D^ E'D^ 
{the homologous sides of similar A are proportional). 

Hence the homologous sides of the polygons are proportional. 

Therefore the polygons are similar, § 319 

{having their homologous A equal, and their homologous sides proportional). 

Q.E.O. 



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154 



PLANE GEOMETRY. — BOOK III. 



Proposition XIII. Theorem. 

332. If two polygons are shrvilar, they are composed 
of the same number of triangles, similar ea^h to each, 
and similarly pUuced. 





B ~C B' & 

Let the polygons ABODE And A'BfaD'E' be similar. 

From two homologous vertices, as E and £?, draw diagonals 
EB, EC, and E'B\ E'QK 

To prove A EAB, EBC, ECD 

similar respectively to A E^A^B\ E^BC\ E^O^U. 

Proof. In the A EAB and E^A^B\ 

ZA = ZA\ §819 

(being homologous A of similar polygons) ; 
AE AB 



and 



319 



A'E' A'B 
{being homologous sides of similar polygons). 
.'. A EAB and E'A^B' are similar, § 326 

{having an /.of the one equal to an Z of the other, and the including sides 
proportional). 

Also, Z ABO= Z A'B'C\ (1) 

{being homologous A of similar polygons). 
And Z ABE= Z A'B'E\ (2) 

(being homologous A of similar A). 
Subtract (2) from (1), 

ZEBC=ZE'B'C', Ax. 3 



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Now 



And 



SIMILAR POLYGONS. 155 

EB _ AB 

(being homologous sides of similar ^). 
BC AB 



BO' A'B 
(being homologous sides of similar polygons), 

EB BC 



Ax. 1 



E'B' B'O' 

/. A EBC and E'B'Cf* are similar, § 326 

(hawing an /.of the one equal to an Z of the other, and the including tides 
proportional). 

In like manner we may prove A ECD and E'C'D' similar. 

/ Q.E.D. 

Peoposition XIV. Theoeem. 

333. The perimeters of two similar polygons have 
the sam^e ratio a^ any two how^ologous sides. 
E 





B c B' a . 

Let the two similar polygons be ABODE and AfB^CfD'E', 
and let P a^nd F* represent their perimeters. 

To prove P:P = AB: A'B\ 

AB : A'B' = BC: B'C = CB : C'U, etc., § 319 
{)he homologous sides of similar polygons are proportional). 

.-. AB+BC, etc. : A'B' + BC\ etc. = ^5 : A'B\ § 303 
iyn a series of equal ratios the sum of the antecedents is to the mm of the 
consequents as any antecedent is to its consequent), 

Thatis, P:P = AB:A'B'. <^eo. 



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156 



PLANE GEOMETRY. — BOOK III. 



Numerical Properties of Figures. 
Proposition XV. Theorem. 

Ni< 334. If in a right triangle a perpendicular is drawn 
from the vertex of, the right angle to the hypotenuse: 

I. The perpendicular is a mean proportional he- 
tween the segments of the hypotenuse. 

II. Each leg of the right triangle is a mean pro- 
portional between the hypotenuse and its a^acent 
segment. 




In the right triangle ABC, let BF be drawn from the 
vertex of the right angle B, perpendicular to AC, 

I. To prove AF\ BF=^ BF\ FO, 
Proof. In the rt. A ^^i^and BAC 

the acute Z ^ is common. 
Hence the A are similar. § 323 

In the rt. A J^OFand BOA 

the acute Z (7 is common. 
Hence the A are similar. . § 323 

Now as the rt. A ^^JPand CBFqxq both similar to ABC, 
they are similar to each other. 
In the similar A ^^i^and CBF, 

AF^ the shortest side of the one, 
BF, the shortest side of the other, 
BF, the medium side of the one, 
FC, the medium side of the other. 

II. To prove AC: AB = AB : AF, 
and AC. BO = BO '.Fa 



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NUMERICAL PROPERTIES OP FIGURES. 167 

In the similar A ABC a.nd AjBF, 

AC, the longest side of the one, 
: ABy the longest side of the other, 
: : AB, the shortest side of the one, 
: AF, the shortest side of the other. 
Also in the similar A ABC dmi. FBC, 

AC, the longest side of the one, 
; BC, the longest side of the other, 
BC, the medium side of the one, 
FC, the medium side of the other. q. k. o. 

336. CoR. 1. The squares of the two legs of a right triangle 
are proportional to the adjacent segments of the hypotenuse. 
The proportions in II. give, by § 295, 

AS=^ACy.AF, and B^-=^ACy.CF. 
By dividing one by the other, we have 
JS _ ACy.AF _AF 
~BC^ ACxCF CF' 

336. Cor. 2. The squares of the hypotenuse and either leg 
are proportional to the hypotenuse and the adjacent segment, 

p AG" _^ ACxAC _AC 

°^ A^ ACxAF AF 

337. Cor. 3. An angle inscribed in a semicircle is a right 
angle (§ 264). Therefore, 

I. The perpendicular from any point in 
the circumference to the diameter of a circle 
is a m£an proportional between the segments ] 
of the diamete7\ 

II. The chord dravmfrom the point to either extremity of the 
diamete7' is a mean proportional between the diameter and the 
adjacent segment. 

Remabe. The pairs of corresponding sides in similar triangles may be 
called longest, shortest, medium, to enable the beginner to see quickly 
these pairs ; but he must not forget that two sides are homologous, not 
because they appear to be the longest or the shortest sides, but because 
they lie opposite corresponding e^ual angles. 




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158 



PLANE GEOMETRY. — BOOK III. 



Peoposition XVI. Theorem. 

^v^ 338. The sum of the squares of the two legs of aright 
triangle is equal to the square of the hypotenuse* 
B 




Let ABC be a right triangle with its right angle at B, 

To prove AF + BO" = AC\ 

Proof. Draw BFl. to AO, 

Then IF = ACxAF § 334 

and W = ACx CF 

By adding. AF + BC^ = A C{AF+ CF) = AC\ a e. d. 

339. Cor. The sqiuire of either leg of a right triangle is equal 
to the difference of the squares of the hypotenuse and the other leg, 

340. Scholium. The ratio of the diagonal of a 
square to the side is the incommensurable num- 
ber V2. For ii AC IS the diagonal of the square 
ABCD, then 

AO'=-AF + SG\or AC'^^AF. 




Divide by AF, we have 



AC' 



AC 
or -T^=-V2. 



IF AB 

Since the square root of 2 is incommensurable, the diagonal 
and side of a square are two incommensurable lines. 

341. The projection of a line CD upon a straight line AB is 
that part of the line AB comprised q 

between the perpendiculars CP and 
DR let fall from the extremities of 
CD. Thus, PB is the projection of 
CD upon AB. ^ 



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NUMERICAL PROPERTIES OF FIGURES. 



159 



UTi 



Peoposition XVII. Theorem. 

842. In any triangle, the square of the side opposite 
n a^cute angle is equal to the sum of the squares of 
the other two sides diminished hy twice the product 
of one of those sides and the projection of the other 
upon that side. 

A 





D 

Fig. 1. 

Let C be an acute angle of the triangle ABC, and 
DC the projection of AC upon BC. 

To prove AW =-£0" + 10" ~2BCx DG, 

Proof. If D fall upon the base (Fig. 1), 

DB = BC-DC] 

If D fall upon the base produced (Fig. 2), 

DB = DC-Ba 
In either case, 

D^^BC' + DC^-r-2BCxDa 

Add AD to both sides of this equality, and we liave 

A^ + D^ = BC^ +A^ +DC^-2BCx DC. 

But AD^+D^=^A^, §338 

and lD^+DO^ = AO^, 

(ihe turn of the $quares of the two legs of art. A is equal to the square 
of the hypotervuse). 

Put ^5* and AC^ for their equals in the above equality, 

J]^ = W + A0'-'2B0x DC, 

a E. D. 



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160 



PLANE &EOMBTRY, — BOOK IH. 



Proposition XVIII. Thbobbm. 

843. In any obtuse triangle, the square of the side 
opposite the obtuse angle is equal to the sum of the 
squares of the other two sides increased by twice tJ^ 
product of one of those sides and the pryectUm' of 
the other upon tha4> side. 
A 




Let C be the obtuse angle of the triangle ABC, and 
CD be the projection of AC upon BC produced. 

Toprove 1^ = BC^ + AC* + 2BCx DO. 

Proot D£^BC+DO, 

Squaring, D^ = BC* + BO^ + 2B0x DO. 

Add AD to both sides, and we have 

A^ + D]?-=BC*+AD'+D0'+2B0xD0. 

But AP + D^=:AF, §838 

and AD' + DO* = AC^, 

(the mm of the tquares of the two legs of art ^ii egual to the square 
of the hypotenuse). 

Put AW and AC* for their equals in the above equality, 

JF = BC* + AC* + 2BCX DC 

aE.o. 

Note. The last three theorems enable as to compute the lengths of 
the altitades if the lengths of the three sides of a triangle are known. 



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NUMERICAL PROPERTIES OP FIGURES. . 161 

^ Proposition XIX. Theorem. 

844. I. The sum of the sqiuares of two sides of a tri' 
angle is egiiaZ to twice the square of half the third 
side increased by twice the square of the median upon 
thai; side. 

II. The difference of the squares of two sides of a 
triangle is equal to twice the product of the third 
side by the prqfection of the median upon that side, 

A 



« MD 

In the triangle ABC let AM he the median, and MD 
the projection of AM upon the side BO. Also let AB 
be greater than AG, 
Toprove I. U^ + AO" = 2BW + 2AM', 

11. AI?-A0'^2£CxMD. 
Proof. Since AB>AC, the Z. AMB will be obtuse, and 
the Z AMO will be acute. § 152 

Then Aff = BM^ + X¥' + 2BMx MD, § 343 

{in any ohttLse A the square of the side opposite the obtuse Z is equal to the 
sum of the squares of the other two staes increased by twice the product 
of one of those sides and the projection of the other on that side) ; 

and AC^ = MC^ + AM*-2MCxMD, §342 

(i7i amy A the square of the side opposite an acute Z is equal to the sum of 
the squares of the other two sides diminished by twice the product of one 
of those sides and the projection of the other upon that side). 

Add these two equalities, and observe that BM= MO, 
Then AB' + AC^ = 2BM^ + 2AM'. 

Subtract the second equality from the first. 
Then AB'- AC^ = 2B0x MD. ^ ^^o. 

Note. This theorem enables us to compute the lengths of the medians 
if the lengths of the three sides of the triangle are known. 



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162 PLANE GEOMETRY. — BOOK III. 



Proposition XX. Theorem. 

345. If any chard is drawn through a -fixed point 
within a circle, the product of its segments is con- 
stant in whatever direction the chord is drawn. 




Let a,ny two chords AS and CD intersect at 0. 
To prove OAxOB=ODx 00. 

Proof. Draw ^Cand BD. 

In the A AOO B,ni BOD, 

AO^jLB, §268 

{fMch, being Tneasured by J arc AD). 

ZA = ZI), §263 

(each being measured by J arc BO). 

.'. the A are similar, § 322 

{two kk are nmilar when two A of the one are equal to two A of the other). 
Whence OA, the longest side of the one, 
: OD, the longest side of the other, 
00, the shortest side of the one, 
OB, the shortest side of the other. 

.-. OAxOB = ODxOO §295 

ai.D. 
346. Scholium. This proportion may be written 

OA^oq gA_ 1 

OB OB' OB^QB' 
00 
that is, the ratio of two corresponding segments is equal to 
the reciprocal of the ratio of the other two corresponding 
segments. In this case the segments are said to be reciproeaUy 
proportional. 



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NUMERICAL PROPERTIES OF FIGURES. 163 



\ Proposition XXI. Theorem. 

847. If frcfm a fixed point ivithout a circle a secant 
is drawn, the product of the secant and its external 
segment is constant in whatever direction the secaM 
is drawn. 




-3 

Let OA and OB be two secants drawn from point 0. 
Topr(yve OA X 00= OB X OD. 

Proof. Draw BC and AD. 

In the A OAD and OBO 

Z is common, 

ZA=/.B, §263 

(each being measured by J arc CD). 

.'. the two A are similar, § 322 

(two ^ are similar when two A of the one are equal to two A of the other). 

Whence 0-4, the longest side of the one, 

: OBy the longest side of the other, 

: OD, the shortest side of the one, 

; 00, the shortest side of the other. 

.-. OA X 00= OB X OB. § 295 

aE.D. 

Remabe. The above proportion continues true if the secant OB turns 
about until B and D approach each other indefinitely. Therefore, by 
the theory o f lim its, it is true when B and D coincide at H. Whence, 
Oil X 00- Off*. 

This truth is demonstrated diredtly in tht next theorem. 



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164 PLANE aEOMETRY. — BOOK III. 



Proposition XXII. Theorem. 

848< If from a point luithout a circle a seccmt and 
a tangent are drawn, the tangent is a mean propor- 
Hanoi betwem the whole secant oind the external 
segm^ent. 




Let OS he a tangent and 00 a secant drawn from 
the point to the circle MBC. 

To prove 00\OB^OB\OM. 

Proof. Draw BMzxA BQ. 

In the A O^Jf and OBO 

Z is common. 

Z OBM is measured by \ arc MB, § 269 

(being an Z formed by a tangent and a chord), 

Z (7is measured by ^ arc BM, § 263 

{being an inscribed Z), 

.\jiOBM=Z.G. 

.'. A 05(7 and OBMbx^ similar, § 322 

{having two A of the one equal to two A of the other). 

Whence OC, the longest side of the one, 

: OB, the longest side of the other, 

: : OB, the shortest side of the one, 

: OMf the shortest side of the other. 

Q.i.a 



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NUMERICAL PROPERTIES OF FIGURES. 165 

Proposition XXIII. Theorem. 

849. The square of the bisector of an angle of a 
triangle is equal to the prodioct of the sides of this 
angle diminished by the product of the segments 
determined by the bisector upon the third side of the 
triangle* 



Let AD bisect the angle BAC of the triangle ABO. 

To prove Alf = ABx AC- DB x DC, 

Proof. Circumscribe the O ABQ about the A ABC. § 285 

Produce AD to meet the circumference in E, and draw EG, 

Then in the A ABD and AEC, 

ABAD = Z.CAE, Hyp. 

AB-^AE, §263 

(tach. being measured by J the arc AC). 

.-. A ABB and AECsire similar, § 322 

(two Ai are gimUar if two A of the one are equal respectively to two A of 

the other). 

Whence AB^ the longest side of the one, 

AE, the longest side of the other, 

; AD, the shortest side of the one, 

; ACy the shortest side of the other. 

.\ABxAC=ADxAE §295 

^AD(,AD + DE) 

=^AP + ADxDE. 

But AD X DE= DB x DC, § 345 

(the product of tTic segments of a chord drawn through a fixed point in 
aOis constant). 

.\ABxAC=:AD' + DBxDa 

Whence ZD' = ABx AC- DBx DC. ^^ o. 

Note. This theorem enables ns to compute the lengths of the bisectors 
of tht angles of a triangle if the lengths of the sides are known. 



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166 PLANE GEOMETRY. — r BOOK III. 



Proposition XXIV. Theorem. 

850. In any triangle the product of two sides is 
equal to the product of the diameter of the circum- 
scribed drdle by the altitude upon the third side. 




Let ABC be a triaaigle, AD the altitude, and ABO 
the circle circumscribed about the triangle ABC. 

Draw the diameter AU, and draw jEO. 

To prove ABxAO= AE X AD. 

Proof. In the A ABD and AEC, 

Z BDA is a rt. Z, Oons. 

Z^OAisart.Z, §264 

{being inscribed in a semicircle), 

Q.r\d ZB = ZE. §263 

.-. A ABD and AEOsire similar, § 823 

(two rt. ^ having an acute Z. of the one equal to an acute Z of t?ie other 
are similar). 

Whence AB, the longest side of the one, 
: AE, the longest side of the other, 
: : ADj the shortest side of the one, 
: ACf the shortest side of the other. 

.:. AB X A0= AEx AD. § 296 

a CD. 

Note. This theorem enables us to compute the length of the radius of 
a circle circumscribed about a triangle, if the lengths of the three BidM 
of the triangle are known. 



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problems of construction. 167 

Problems of Construction. 

Proposition XXV. Problem. 

351. To divide a given straight line into parts pro- 
portional to any number of given lines. 

H K n 




Let AB, m, n, and p, be given straight lines. 

To divide AB into parts proportional to m, w, andp, 

Ckmstrnotion. Draw AX, making an acute Z with AB. 

On ^Xtake ^(7=m, CE=n, EX^p. 

Draw BX. 

From E and O draw EK and OH II to BX. 

JE'and JSTare the division points required. 

T. . fAK\ AH HK KB « oaq 

^*' \jEr'Ad--CE'-EX' §^°^ 

(a lim dravm through two sides 0/ a A il to the third side divides thou 
sides proportionally). 

/. AH : HK :KB = AC :CE : EX. 

Substitute m, n, and p for their equals AC, CE, and EX. 

Then AH : HK : KB == m : n : o. 

aE.F. 



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168 PLANE GEOMETRY. — BOOK III. 



Proposition XXVI. Problem. 

352. To find a fourth proportional to three given 
straight lines. 



m 




Let the three given lines be m, n, andp. 

To find a fourth proportional to m, w, andp. 

Draw Ax and At/ containing any acute angle. 

OonBtruction. On Ax take AB equal to m, £G=^n. 

On Ay take AD=p, 

Draw BI). 

From Cdraw OF \\ to BB, to meet Ay at F. 

DF is the fourth proportional required. 

Proof. AB : BC==AI>: DF, § 309 

(a line drawn through two sides of a AW to the third side divides those 
sides proportionally). 

Substitute m, 7i, andp for their equals AB, BO, and AD. 

Then m : n =p : DF, 

aE. F. 



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PEOBLEMS OF CONSTRUCTION. 169 



Pboposition XXVH. Problem. 



868. To find 


a 


third 


proportional to 


two 


given 


straight lines. 


A 

t 


\ 


m 


•t 




4 










\ 
\ 


n 







! 


\ 
\ 

\ 






D^ 







— .\ij 







Let m and n be the two given straight lines. 

To find a third proportional to m and n. 

Oonstmotion. Construct any acute angle A, 

and take AB^m, AC—n. 

Produce AB to D, making BD == AO. 

Join Ba 

Through D draw DE II to jB(7to meet reproduced at E. 

CE is the third proportional to -4-B and AC, 

Proof. AB : BI) = AC: CE. § 309 

(a line drawn through txoo sides of a AW to the third side divides those 
sides proportionally). 

Substitute, in the above proportion, ^Cfor its equal BD. 

Then AB:AC=AC: CE, 

That is, m : n = n : CE. 

Q.E.F. 



Ex. 217. Confltruct », if (1) a; - ^, (2) a? - ^. 

c c 

Special Cases : (1) a = 2, 6 = 3, c = 4; (2)a-3, ft = 7, c-11; (3) 

a-2, c-3; (4)a-3, c = 5; (5)a = 2c. 



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170 PLANE GEOMETRY. — BOOK IH. 

Proposition XXVIII. Problem. 

364. To -find a mean propofrtional between two given 
gtraigkb lines. 





/ 


'•'i . 










/ 








\ 




m 




1 








\ 






i .• 


"m" 


"c 


n 


B 


'E 


n 


I / 





Let the two given lines bem and n. 

To find a mean proportional between m and n. 
CkmstrnctioiL On the straight line AE 

take AO^m, and CB = n. 
On AB as a diameter describe a semi-circumference. 
At C erect the -L OS to meet the circumference at S, 
CH'iB a mean proportional between m and n. 

Proof. .-. AC :CH=CH : CB, § 837 

(the X let fall from a point in a drcurnference to the diameter of a circle 
is a mean proportional between the segments of the diameter). 

Substitute for -4(7 and CB their equals m and n. 

Then m : Cff ^ CH : n. 

355. A straight line is said to be divided in extreme and 
mean ratio, when the whole line is to the greater segment as 
the greater segment is to the less. 



Ex. 218. Construct as if as =\/a6. 

Special Cases : (l)a-2, 6 = 3; (2)o = l, d-6; (3)a«3, 6-7. 



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PROBLEMS OF CONSTRUCTION. 171 



>. . Proposition XXIX. Problem. 

356. To divide a given line in extreme and mean 



raMo. 






\0 






0^-- 



A C B 

LetAB be the given line. 

To divide AB in extreme and mean ratio. 
Oonstrnotion. At B erect a -L BE equal to one-half of AB. 
From -27 as a centre, with a radius equal to EB^ describe a O. 
Draw AE, meeting the circumference in -P'and O, 
OnAB\^\LQAC=AF. 
On BA produced take AC^ ^ AG. 
Then AB is divided internally at C and externally at (7' 
in extreme and mean ratio. 

Proof. AG'.AB=^AB:AF, §348 

ijf from a point without aO a secant and a tangent are drawn, the tan- 
gent is a mean proportional between the whole secant and the external 
segment). 

Then by § 301 and § 300, 

AG-'AB:AB=AB-AF:AF, (1) 

AG+AB:AG = AB + AF : AB. (2) 

By construction FG = 2 EB = AB. 

.'. AG-AB=^AG-FG = AF=Aa 
Hence (1) becomes 

AO:AB=BC:AC; 
or, by inversion, AB : A0= AO : BO. § 299 

Again , since 0'A = AG = AB+ AF, 
(2) becomes O^B : C'A = O'A : AB. 

Q.CP. 



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172 



PLANE GEOMETRY. — BOOK III. 



Proposition XXX. Problem. 

867. Upon a given line homologous to a given side 
of a given polygon, to construct a polygon similar to 
the given polygon. 
E 





Let A*E' be the given line homologous to AE of the 
given polygon ABODE, 

To construct on A^Ef a 'polygon similar to the given polygon. 

Oonstruction. From E draw the diagonals EB and EC. 

From E' draw E^B\ E'G\ and E^D\ 

making ^^'^'J5', B^E^C\ and C^E^D' equal respectively to 

A AEB, BEC, and CED. 

From A' draw A*B\ making Z E'A'B= /. EAB, 

and meeting E^B^ at BK 
From B draw B^C\ making /L E'B'Q^ = Z EBO, 

and meeting E^C at C. 

From 0' draw C^jy, making Z E'C'B' = Z ^OZ), 

and meeting E*J)' at i)'. 

Then A'B'C^B'E' is the required polygon. 

Proof. The corresponding A ABE a,ni A'B'E\ EBO and 
E'B'C\> ECD and E'C'D' are similar, § 322 

{two ki. are similar if they have two A of the one equal respectively to two 
A of the other). 

Then the two polygons are similar, § 331 

{two polygons^composed of the same number of ^ similar to each other and 
similarly placed, are svmilar). 



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PEOBLEMS OF COMPUTATION. 



173 



Problems op Computation. 
219. To compnte the altitudes of a triangle in terms of its sides. 

o c 





Fig. 1. 



c F D 
Fio. 2. 



At least one of the angles ^ or ^ is acnte. Suppose it is the angle B. 

§338 
§342 



In the A CDB, 
In the A ABC, 

Whence, 



h^^a'-BD'. 
b'^a^^c»-2cxBD. 



BD^ 



2c 



Hence A-^^a2-.(^' + ^-^y ^^^'^-(a'-^^-^>7 
4c» 4c2 

(2ac -^a^ + c^- h^){2ac -a^-c^ + h") 

4c» 

4c« 
^ (g + 6 + c) (g + c - ft) (6 + g — c) (6 - g -h c) 
4c« 
g + 6 + c = 2s. 
a + c-6 = 2(8-6), 
6 + g — c= 2(5 — c), 
6 — g + c = 2(s — g). 
,a_ 28 X 2(8 - g) X 2(8 -h)x 2{6 - c) 
4c» 
By simplifying, and extracting the square root, 

;i = ?V8(8-g)(8-6)(8-c). 
c 

220. To compute Uie medians of a triangle in terms of its sides. 

By i 344, g» + 6» - 2m« + 2 ^| Y- (Fig. 2) 



Let 
Then 



Hence 



Whence 



4m2 = 2(g« + 6«)-.c«. 
.m=iV2(g» + 6»)-c». 



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174 



PLANE GEOMETRY. BOOK III. 



221. To compute the bisectors of a triangle in terms of the sides. 
By §349, f^=^ah'~ADxBD. 

AD^BD _ AD + BD _^ c 
h a a + 6 a+6 

he ^. Dn_ ac 



By J 313, ^=£i^ = 



.i4i>- 



Whtnce 



a + 6 



and 5i>- 



a + h 



ah(? 



Whence 



(a + hf 

(a + hf 
o6(a + 5-f-c)(a-f6~c) 

o6x2gx2(g — c) 

(a + 6)« 

i = -y/ahi U — c). 

a+6 




222. To compute the radius of the circle circumscribed about a tri- 
angle in terms of the sides of the triangle. ^ *^v^^ 

By? 350, ABxAC^AExAD, 
or hc^2IixAD. 



But 
Whence 



AD^^V8(8-a){8- 
a 

R = ^he 



b)(s-c). 




4Vs(8-a)(«-6)(8-c) 

223. If the sides of a triangle are 3, 4, and 5, is the angle opposite 5 
right, acute, or obtuse ? 

224. If the sides of a triangle are 7, 9, and 12, is the angle opposite 
12 right, acute, or obtuse ? 

225. If the sides of a triangle are 7, 9, and 11, is the angle opposite 
11 right, acute, or obtuse ? 

226. The legs of a right triangle are 8 inches and 12 inches ; find the 
lengths of the projections of these legs upon the hypotenuse, and the dis- 
tance of the vertex of the right angle from the hypotenuse. 

^- 227. If the sides of a triangle are 6 inches, 9 inches, and 12 inches, 
find the lengths (1) of the altitudes ; (2) of the medians ; (3) of the bisec- 
tors ; (4) of the radius of the circumscribed circle. 



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EXERCISES. 176 



Theorems. 

228. Any two altitudes of a triangle are inversely proportional to 
the corresponding bases. 

229. Two circles touch at P. Through P three lines are drawn, meet- 
ing one circle in ^, -6, C, and the other in A\ B\ (7, respectively. Prove 
that the triangles ABC^ A'B'Cy are similar. 

^230. Two chords AB, CD intersect at M, and A is the middle point of 
the arc CD. Prove that the product AB x ^il/ remains the same if the 
chord AB is made to turn about the fixed point A, 

Hint. Draw the diameter AE^ join BE,\f^ compare the triangles 
thus formed. 

231. The sum of the squares of the segments of two perpendicular 
chords is equal to the square of the diameter of the circle. 

Hint. If AB, CD are the chords, draw the diameter BE, join AC, 
ED, BD, and prove that AC = ED. Apply g 338. 

232. In a parallelogram ABCD, a lino DE is drawn, meeting the 
diagonal AC in F, the side BC in G, and the side AB produced in E. 
?roYeih&i DF'^FGxFE. 

233. The tangents to two intersecting circles drawn from any point 
in their common chord produced, are equal. (§ 348.) 

234. The common chord of two intersecting circles, if produced, will 
bisect their common tangents. (J 348.) 

y 235. If two circles touch each other, their common tangent is a mean 
proportional between their diameters. 

Hint. Let AB be the common tangent. Draw the diameters AC, BD. 
Join the point of contact Pto A,B, C, and D. Show that APD and BPC 
are straight lines ± to each other, and compare A ABC, ABD. 

"7 236. If three circles intersect one another, the common chords all pass 
-^rough the same point. , 

Hint. Let two of the chords AB and CD 
meet at 0. Join the point of intersection E 
to 0, and suppose that EO produced meets 
the same two circles at two different points P 
and Q. Then prove that OP=OQ\ hence, 
tb»t the points P and Q coincide, 




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176 



PLANE GEOMETRY. — BOOK Til. 



237. If two circles are tangent internally, all chords of the greater 
circle drawn from the point of contact are divided proportionally by the 
circumference of the smaller circle. 

Hint. Draw any two of the chords, join the points where they meet 
the circumferences, and prove that the A thus formed are similar. 

' 238. In an inscribed quadrilateral, the product of the diagonals is 
equ&l to the sum of the products of the opposite sides. 

Hint. Draw Z)-E;, making ZCD-E=Z -42)5. The A 
^ ABD and CDE are similar. Also the ^ BCD and 
ADE are similar. 

~ 239. The sum of the squares of the four sides of 
any quadrilateral is equal to the sum of the squares 
of the diagonals, increased by fourstimes the square 
of the line joining the middle points of the diagonals. 
Hint. Join the middle points F, E, of the diag- 
onals. Draw EB and ED. Apply § 344 to the 
^ ABC and ADC, add the results, and eliminate 
BE^ + DE"^ by applying § 343 to the 4 BDE. 





240. The square of the bisector orari exterior angle of a triangle is 
equal to the product of the ex^rnal segments deter- 
mined by the bisector upon one of the sides, dimin- 
ished by the product of the other two sides. 

Hint. Let CD bisect the exterior /. BCE of 
the A ABC Circumscribe a G about the A. pro- 
duce DC to meet the circumference in F, and draw BF. 
BCF similar. Apply J 347. 

^ — J 241. If a point is joined to the vertices of a triangle ABC, and 
/through any point A^ in OA a line parallel to AB is drawn, meeting OB 
at B\ and then through B^ a line parallel to BG, meeting OC at (7, 
and C is joined to A\ the triangle A'B'C will be similar to the tri- 
angle ABC 



Vrowe ^ACD, 



/ 



242. If the line of centres of two circles meets the circumferences at 
the points A, B, Q D, and meets the common exterior tangent at P, then 
PAxPD='FBxPC 

^ ' 243. The line of centres of two circles meets the common exterior 
tangent at P, and a secant is drawn from P, cutting the circles at the 
consecutive points E, P, G, H. Prove that PExPH=PFx PQ, 



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EXERCISES. 177 



NXTMEBICAL EXEBCISES. 

^ 244. A line is drawn parallel to a side AB of a triangle ABC, and 
^cutting AC in A BC in E. If AD : i>C- 2 : 3, and AB^20 inches, 
^d i>-&. I i^ 

y 245. The sides of a triangle are 9, 12, 15. Find the segments made by 
bisecting the angles. ({ 313.) 

«^246. A tree casts a shadow 90 feet long, when a vertical rod 6 feet 
mgh casts a shadow 4 feet long. How high is the tree ? 

y*j 247. The bases of a trapezoid are represented by a, b, and the altitude 
Wy h. Find the altitudes of the two triangles formed by producing the 
legs till they meet. 

^S^8. The sides of a triangle are 6, 7, 8. In a similar triangle the side 
l^i(mologons to 8 is equal to 40. Find the other two sides. 

'*y%49. The perimeters of two similar polygons are 200 feet and 300 feet. 
If a side of the first polygon is 24 feet, find the homologous side of the 
second polygon. 

^^^^50. How long must a ladder be to reach a window 24 feet high, if 
4ie lower end of the ladder is 10 feet from the side of the house ? 

^ 251. If the side of an equilateral triangle » a, find the altitude. 

252. If the altitude of an equilateral triangle = h, find the side. 

253. Find the lengths of the longest and the shortest chord that can 
le drawn through a point 6 inches from the centre of a circle whose 

radius is equal to 10 inches. 

^ 254. The distance from the centre of a circle to a chord 10 inches long 
is 12 inches. Find the distance from the centre to a chord 24 inches long. 

£5. The radius of a circle is 5 inches. Through a point 3 inches from 
entre a diameter is drawn, and also a chord perpendicular to the 
ster. Find the length of this chord, and the distance from one end 
of the chord to the ends of the diameter. 

266. The radius of a circle is 6 inches. Through a point 10 inches 
Tom the centre tangents are drawn. Find the lengths of the tangents, 
and also of the chord joining the points of contact 

\ 257. If a chord 8 inches long is 3 inches distant from tba centre of 
Ihe circle, find the radius and the distances from the end of the chord to 
the ends of the diameter which bisects the chord. 



;* 



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-^ 



178 PLANE GEOMETRY. — BOOK IH. 

268. The radius of a circle is 13 iDchea. Through a point 5 inches 
fpm the centre any chord is drawn. What is the product of the two seg- 
ments of the chord ? What is the length of the shortest chord that can 
be drawn through the point ? 

;> 259. From the end of a tangent 20 inches long a secant is drawn 
through the centre of the circle. If the exterior segment of this secant 
is 8 inches, find the radius of the circle. 
2C0. The radius of a circle is 9 inches ; the length of a tangent is 12 
iches. Find the length of a secant drawn from the extremity of the 
tangent to the centre of the circle. 

261. The radii of two circles are 8 inches and 3 inches, and the dis- 
tance between their centres is 15 inches. Find the lengths of their com- 
mon tangents. 

•<^62. Find the segments of a line 10 inches long divided in e^eme 
^d mean ratio. ^ 

/263. The sides of a triangle are 4, 5, 6. Is the largest angle acute, 
right, or obtuse ? 

__^ Problems. 

264. To divide one side of a given triangle into segments proportional 
to the adjacent sides. (§ 313.) 

265. To produce a line ^15 to a point Cso that AB : AC=» 3 : 5. 

266. To find in one side of a given triangle a point whose distances 
from the other sides shall be to each other in a given ratio. 

267. Given an obtuse triangle ; to draw a line from the vertex of the 
obtuse angle to the opposite side which shall be a mean proportional 
between the segments of that side. 

268. Through a given point P within a given circle to draw a chord 
AB so ihAtAP:BP^2:S. 

269. To draw through a given point P in the arc subtended by a chord 
AB a chord which shall be bisected by AB. 

270. To draw through a point P, exterior to a given circle, a secant 
FAB so that P.4: ^^- 4: 3. 

271. To draw through a point P, exterior to a given circle, a secant 
PAB so that Iff ^PAx PB. 

272. To find a point P in the arc subtended by a given chord AB so 
thatPil:P£»3:l. 



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EXEKCISES. 179 

' 273. To draw through one of the points of intersection of two circles 
i, secant so that the two chords that are formed shall he to each other 
in the ratio of 3 : 5. 

'274. To divide a line into three parts proportional to 2, f , J. 

275. Having given the greater segment of a line divided in extreme 
and mean ratio, to construct the line. 

276. To construct a circle which shall pass through two given points 
and touch a given straight line. 

' 277. To construct a circle which shall pass through a given point and 
touch two given straight lines. 

278. To inscribe a square in a semicircle. 

279. To inscribe a square in a given triangle. 

Hint. Suppose the problem solved, and DEFG the inscribed square. 
Draw CM 11 to AB, and let AF produced meet 
CM in M. Draw Off and MN ±io AB, and 
produce AB to meet MN at N. The A ACM, 
AOF are similar; also the A AMN, AFE €k 
are similar. By these triangles show that A 
the figure CMNH is a square. By construct- /A 
ing this square, the point F can be found. ATl 

/ 280. To inscribe in a given triangle a rectangle similar to a given 
rectangle. 

281. To inscribe in a circle a triangle similar to a given triangle. 

282. To inscribe in a given semicircle a rectangle similar to a given 
rectangle. 

283. To circumscribe about a circle a triangle similar to a given 
triangle. 

284. To construct the expression, x - ^^ ; that is, — x -• 

de d e 

285. To construct two straight lines, having given their sum and 
their ratio. . ^ 

286. To construct two straight lines, having given their difference 
and their ratio. jm 

287. Having given two circles, with cpllfe O and Cy, and a point A 
in their plane, to draw through the point A a straight line, meeting the 
circumferences at S&nd Q so that AB :AC'*l:2. 

Hint. Suppose the problem solved, join OA and produce it to D, 
making OA.AD^l-.Z Join DC; ^OA B, ADO are similar. 



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BOOK IV. 



AREAS OF POLYGONS. 

858. The area of a surface is the numerical measfu/re of the 
surface referred to the unit of surface. 

The unit of surface is a square whose side is a unit of length; 
as the square inch, the square foot, etc. 

369. Equivalent figures are figures having equal areas. 

Proposition I. Theorem. 

860. The areas of two rectangles having equal alti- 
tudes are to each other as their bases. 



J_ 




-5 — ' ■ • ■ ^^ - — 5 ^ 

Let the two rectSLngles be AC and AF^ having the 
same altitude AD. 

rp TQQi.AO AB 

To prove — " . ■„ = --r=i' 

^ rect.^i^ AE 

Proof. Case I. When AB and AE are commensurable. 

Suppose AB and AE have a common measure, as ^0, 

which is contained in AB seven times and in AE four times. 

AB _1 

AE 4 

Apply this measure to AB and AE, and at the several 

points of division erect Js. 

The rect. -4 (7 will be divided into seven rectangles, 

and the rect. AF will be divided into four rectangles. 



Then 



(1) 



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AREAS OP POLYGONS. 



181 



These rectangles are all equal. 

rectAO __7 
4 
AB 



Hence 



From (1) and (2) 



rect.J.i^ 
rect. AC 



TQct.AF AE 
Case II. When AB and AE are incommensurable. 

I>i ! 1 ! ! ! \0 l>r- 



§186 
(2) 

Ax. 1 



T 



Divide AB into any number of equal parts, and apply one 
of them to AE as often as it will be contained in AE. 

Since AB and AE are incommensurable, a certain number 
of these parts will extend from ^ to a point K, leaving a 
remainder KE less than one of the parts. 
DT2iW KIT W to EF. 
Since AB and -4^ are commensurable, 
rect.^^ AK 



rect. AO AB 
These ratios continue equal, as the unit of measure is indefi- 
nitely diminished, and approach indefinitely the limiting ratios 

rect.^i^ . AE ,. , 

-— and -—- respectively. 

rect AG AB 



rect. AF AE 



§280 



rect. AO AB 
(if two variables are constantly equal, and each approaches a limit, the 
limits are eqiial). q^ e^ q^ 

861. Cor. The areas of two rectangles having eqiial bases are 
io each other as their altitudes. For AB and AE may be con- 
sidered as the altitudes, AI) and AI) as the bases. 

Note. In propositions relating to areas, the words "rectangle," 
*' triangle," etc., are often used for " area of rectangle," " area of tri- 
angle," etc. 



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182 



PLANE GEOMETKY. — BOOK IV. 



Proposition II. Theorem. 

862. The areas of two rectangles are to ea^h other 
as the products of their bases by their altitudes. 




b b' b 

Let R and B' be two rectangles, having for their 
bases b and b\ and for their altitudes a and a'. 
rrr B aXb 

Proof. Construct the rectangle 8, with its base the same as 
that of B, and its altitude the same as that of B'. 

R a 



Then 



8 



§361 



and 



{rectangles having equal bases are to each other as their altitudes) 

8_^b^ 

B' &'' 
(rectangles having equal altitudes are to each other as their bases). 

By multiplying these two equalities, 
B _ aXb 
K a! X V 



§360 



aE.ix 



Ex. 288. Find the ratio of a rectangular lawn 72 yards by 49 yards 
to a grass turf 18 inches by 14 inches. 

Ex. 289. Find the ratio of a rectangular courtyard 18 J yards by 15J 
yards to a flagstone 31 inches by 18 inches. 

Ex. 290. A square and a rectangle have the same perimeter, 100 yards. 
The length of the rectangle is 4 times its breadth. Compare their areas. 

Ex. 291. On a certain map the linear scale is 1 inch to 5 miles. How 
many acres are represented on this map by a square the perimeter of 
which is 1 inch ? 



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AREAS OP POLYGONS. 



183 



Proposition III. Theorem. 

S. The area of a rectangle is equal to the product 
of its ba^e and altitude. 




■0 



Let B he the rectangle, h the base, and a the altl- 
tnde; and let U be a square whose side is equal to 
the linear unit 

To prove the area of JR = axb. 

R axb 



= axb, 



§362 



But 



: the area of B. 



§358 



U 1x1 
(two rectangles are to each other as th^ product of their bases and altitudes). 

U' 
.*. the area of B = a X b. a e, o. 

864. Scholium. When the base and altitude each contain 
the linear unit an integral number of times, this proposition is 
rendered evident by dividing the figure into squares, each 



equal to the unit of measure. Thus, if the base contain seven 
linear units, and the altitude four, the figure may be divided 
into twenty-eight squares, each equal to the unit of measure ; 
and the area of the' figure equals 7x4 units of surface. 



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184 



PLANE GEOMETRY. — BOOK IV. 



(y 



Proposition IV. Theorem. 



866. The area of a parallelogram is equal to the 
product of Us base and altitude, 
B E C F 




A h D A b D 

Let AEFD be a psLrsdlelogram, AD its base, and CD 
its altitude. 

To prove the area of the O AEFD = AB X CD, 
Proof. From A draw AB \i io DC io meet i^^ produced. 
Then the figure ABCD will be a rectangle, with the same 
base and altitude as the O AEFD. 
In the rt. A ABE and DCF 

AB = CD and AE= DF, § 179 

{being opposite sides of a CJ). 

.'.AABE=ADCF, §161 

{two rt. A are equal when the hypotenuse and a side of the one are equal 
respectively to the hypotenuse and a side of the other). 

Take away the A DCF, and we have left the rect. ABCD. 

Take away the A ABE, and we have left the O AEFD. 

.-. rect. ABCD =c= O AEFD. Ax. 3 

But the area of the rect. ABCD ==axb, § 363 

.-. the area of the O AEFD ^axb. Ax. 1 

aE.D. 

366. CoR. 1. Parallelograms having equal bases and equal 
altitudes are equivalent. 

367. CoR. 2. Parallelograms having equal bases are to each 
other as their altitudes ; parallelograms having equal altitudes 
are to each other as their bases ; any two parallelograms are 
to each other as the products of their bases by their altitudes. 



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AREAS OP POLYGONS. 185 



Proposition V. Theorem. 

368. The area of a triangle is equal to one-half 
tlie product of its base by its altitude. 




Let ABC be & triangle, AB its base, and DC its 
altitude. 

To prove the area of the A AJBC= i AB X BC. 

Proof. From C draw CH II to BA. 

From A draw AH II to BO, 

The figure ABCH'is a parallelogram, § 168 

{having its opposite sides parallel), 

and AC\^ its diagonal. 

.'.AABC-=AAIia §178 

{tJie diagonal ofaCJ divides it into two equal A). 

The area of the O ABCH is equal to the product of its 
base by its altitude. § 365 

Therefore the area of one-half the O, that is, the area of 
the A ABC^ is equal to one-half the product of its base by its 
altitude. 

Hence, the area of the A ABO= ^AB X DC, 

Q.E.D. 

869. Cor. 1. Tiai^gles having equal bases and eqital alii- 
iudes are equivalent, 

870. Cor. 2. Triangles having equal bases are to each other 
as their altitudes ; triangles having equal altitudes are io each 
other as their bases ; any two triangles are to each other as the 
produ^cts of their bases by their altitudes. 



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186 



PLANE GEOMETRY. — BOOK IV, 



Proposition VI. Theorem. 



[y 



371. The area of a trapezoid is equal to one^hoHf 
the sum of the parallel sides multiplied by the alti- 
tude. TT T* J)' 




E 



^^ 



y 




A F b D 

Let ABCH he a trapezoid, and EF the altitude. 
To prove area of ABCH = \ {ffC+ AB) EF 
Proof. Draw the diagonal AC, 

Then the area of the A ABC= J {AB X EF), § 368 

and the area of the A AHC^ \ \hC X EF\ 
By adding, area of ABCH= i (AB + EC) EF, a e. o. 

372. Cor. The area of a trapezoid is equal to the product 
of the median hy the altitude. For, by § 191, OP is equal to 
i{E:C+AB)\ and hence 

the area of ABCB:= OP X EF, 

373. Scholium. The area of an irregular polygon may be 
found by dividing the poly- 
gon into triangles, and by 
finding the area of each of 
these triangles separately. 
But the method generally 
employed in practice is to 
draw the longest diagonal, 
and to let fall perpendiculars upon this diagonal from the 
other angular points of the polygon. 

The polygon is thus divided into right triangles and trape- 
zoids ; the sum of the areas of these figures will be the^area 
of the polygon. 




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AREAS OF POLYGONS. 187 



Proposition VII. Theorem. 

874. The areas of two triangles which have an angle 

of the one equal to an angle of the other are to each 

other as the products of the sides including the equal 

angles. 

A 




A ADE AD X AE 


Draw BE. 


A ABC 


AO 


A ABE 


'Al! 


A ABE 


_AB 



Let the triangles ABC and ADE have the common 
angle A. 

rp A ABC ABxAC 

To prove — 

Proof. 

Now ^^^^7 = ^^7^ ^ 

and ^^^^ = ^^, ^370 

A ADE AD ^ 

. (^ having the same altitade are to each other oa their bases). 
By multiplying these equalities, 

AABC _ ABxAO 

A ADE AD X AE 

aE.o. 

Ex. 292. The areas of two triangles which have an angle of the one 
supplementary to an angle of the other are to each other as the products 
of the sides including the supplementary angles. 



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188 



PLANE GEOMETRY. — BOOK IV. 



Comparison of Polygons. 

Proposition VIII. Theorem. 

375. The arecus of two similar triangles are to each 
other flw the squares of any two homologous sides. 





A B A'- O' 

Let the two triangles be AC B and A'O'B'. 
AACB AS 



To prove 



A A^C'B' JTgf« 



Draw the perpendiculars CO and CO', 
rpj^^^ A ACB _ ABxCO 



AB ^00 



A A'O'B' A'B' X CO' A'B' CO' 



§370 



(two ^ are to each other as the products of their bases by their altititdes). 



But 



AB CO 



§328 



A'B' O'O' 

(the homologous altitudes of similar ^ have the same ratio oa their homolo- 
gous bases). 

CO AB 

Substitute, in the above equality, for — — its equal ; 

CO A Jlj 



then 



AW 



AACB 
AA'0*B' A'B^'^A'B' JHS^ 



AB ^ AB 



aE.a 



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C0HPABI80R OF FOLYOONS. 



189 



Proposition IX. Theobem. 

876. The areas of two similar polygons are to each 
other as the squares of any two homologous sides. 
E 





DC B' C" 

Let 3 and 8^ denote the areas of the two similax 
polygons ABC etc., and A'^C etc. 
Toprove S : S' = AB' : ATB^. 

Proof. By drawing all the diagonals from the homologous 
vertices U and ^', the two similar polygons are divided into 
triangles similar and similarly placed. , § 332 

• XB' _ AABE _ f£E\ _ A BOH 
VgrgrV AB'C'E' 



A'B>' 



AA'B>E' 
\0'E'V 



A CPE 
A O'B'E'' 



§376 



(timilar A are to each other at the tguara of any two homohgout tide*). 

™ ^ . A ABE A BCE A CDE 
That 18, — — 



AA'B'E* AB'G'E> AC'D'E' 
A ABE+ BCE+ QBE __ A ABE _ Zg* 

A'B'" 



(in a serici of equal ratios the mm of the antecedents is to the sum of the 
' consequents as any antecedent is to its consequent). 

877. Cor. 1. The areas of two similar polygons are to each 
other as the squares of any two homologous lines. 

378. Cor. 2. The homologous sides of two similar polygons 
have the same ratio as the square roots of their areas. 



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190 



PLANE GEOMETRY. — BOOK IV. 



Proposition X. Theorem. 

379. The square described on the hypotenuse of a 
right triangle is equivalent to the sum of the squares 
on the other two sides. 

Let BE, CHy AF, be squares on the tliree isides of the 
right triangle ABC. 

To prove £0^ ^ AS + AQ^. 

Proof. Through A draw AL II to 
CE, and draw AB and FQ. 

Since A BAC, BAG, and CAR 
are rt. A, CAO and BAH are 
straight lines. 

Since BD= BC, being sides of 
the same square, and BA = BFj 
for the same reason, and since 
Z ABD = Z FBC, each being the 
sum of a rt. Z and the Z. ABC, 

the A ABB = A FBC. 

Now the rectangle BL is double the A ABB, 
{having the same base BD, and the same altitude, the distance between the 
\\s AL and BD), 

and the square^i^is double the A FBC, 

{having the same base FB, and the same altitxide, the distance between the 
lis FB and OC). 

Hence the rectangle BL is equivalent to the square AF. 

In like manner, by joining AE and BK, it may be proved 
that the rectangle CL is equivalent to the square CH. 

Therefore the square BE, which is the sum of the rectangles 
BL and CL, is equivalent to the sum of the squares CH and 
AF. aE.D. 

880. Cor. The sqttare on either leg of a right triangle is 
equivalent to the difference of the sqicares on the hypotemise and 
the other leg. 




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COMPAEISON OP POLYGONS. 



191 



Ex. 293. The square constructed upon the sum of two straight lines 
is equivalent to the sum of the squares constructed upon these two lines, 
increased by twice the rectangle of these lines. 

Let AB and BC be the two straight lines, and AC their sum. Con- 
struct the squares ACOK and ABED upon AC&nd 
AB respectively. Prolong BE and DE until they 
meet KO and CG respectively. Then we have the 
square EFOH, with sides each equal to BC. Hence, 
the square ACOK is the sum of the squares ABED D - 
and EFOff, and the rectangles DEHK and BCFE, 
the dimensions of which are equal to AB and BC. 



BC 



K 









E 



H Q 



C 



D 



E 



Ex. 294. The square constructed upon the difference of two straight 
lines is equivalent to the sum of the squares constructed upon these two 
lines, diminished by twice the rectangle of these lines. 

Let AB And AC he the two straight lines, and -BC their difference. 
Construct the square ABFO upon AB, the square jg- g^ 
ACKH ui>on AC, and the square BEDC upon BC{as 
shown in the figure). Prolong ED until it meets AG 
in L. 

The dimensions of the rectangles LEFG and HKDL 
are AB and AC, and the square BCD E is evidently 
the difference between the whole figure and the sum 
of these rectangles ; that is, the square constructed O F 

upon BC is equivalent to the sum of the squares constructed upon AB 
and AC diminished by twice the rectangle oi AB and AC. 

Ex. 295. The difference between the squares constructed upon two 
straight lines is equivalent to the rectangle of the sum and difference of 
these lines. 

Let ABDE and BCGF be the squares constructed upon the two 
straight lines AB and BC. The difference between 
these squares is the polygon ACOFDE, which poly- 
gon, by prolonging CG to H, is seen to be composed of 
the rectangles ACHE and OFDH. Prolong AE and 
CHto Jand ^respectively, making Elsind HK eAch 
equal to BC, and draw IK The rectangles QFDH 
and EHKI are equal. The difference between the 
squares ABDE and BCGF is then equivalent to the 
rectangle ACKI, which has for dimensions AI'^AB-k-BC, and EH 
^AB-BC 



I i 


' 


B 




H 


d 






tr 





C B 



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192 



PLANE GEOMETRY. — BOOK IV. 



Peoblems of Construction. 

Proposition XI. Problem. 

881. To construct a square equivalent to the sum 
of two given squares. 



R 




^K 



A^- 




Let R and R' be two given squares. 

To construct a square equivalent to B' + H. 
Oonstruction. Construct the rt. Z A. 

Take AC equal to a side of jB', 

AB equal to a side of i? ; and draw BO. 

Construct the square /?, having each of its sides equal to BO, 

S is the square required. 

Proof. BO'^AO' + Aff, §379 

(the square on the hypotenuse of a rt. A is equivaUnt to the sum of the 
squares on the two sides). 



.'.S^B' + S. 



aE.F. 



Ex. 296. If the perimeter of a rectangle is 72 feet, and the length is 
equal to twice the width, find the area. 

Ex. 297. How many tiles 9 inches long and 4 inches wide will be 
required to pave a path 8 feet wide surrounding a rectangular court 120 
feet long and 36 feet wide ? 

Ex. 298. The bases of a trapezoid are 16 feet and 10 feet ; each leg 
is equal to 5 feet. Find the area of the trapezoid. 



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PROBLEMS OP CONSTRUCTION. 



193 



Proposition XII. Problem. 

882. To construct a square equivalent to the differ- 
ence of two given squares. 











Bk 










R 




R' 


\ 


> 









Let B be the smaller square and B^ the larger. 

To construct a square equivalent to R^ — R, 
Oonstruction. Construct the rt. Z A. 

Take AB equal to a side of R, 

From .5 as a centre, with a radius equal to a side of R\ 

describe an arc cutting the line -^JTat C. 

Construct the square 8, having each of its sides equal to AC, 

8 is the square required. 

Proof. AC^^^BC^-A^, §380 

(the square on diher leg of a rt. L is equivalent to the difference of the 
squares on the hypotenuse and the other leg), 

.'.S^R'-R. 



Ex. 299. Construct a square equivalent to the sum of two squares 
whose sides are 3 inches and 4 inches. 

Ex. 300. Construct a square equivalent to the difference of two 
8qua|& whose sides are 2^ inches and 2 inches. 

Ex. 301. Find the side of a square equivalent to the sum of two 
•quares whose sides are 24 feet and 32 feet. 

Ex. 302. Find the side of a square equivalent to the difference of two 
squares whose sides are 24 feet and 40 feet. 

Ex. 303. A rhombus contains 100 square feet, and the length of one 
diagonal is 10 feet. Find the length of the other diagonal. 



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194 



PLANE GEOMETRY. — BOOK IV. 



Peoposition XIII. Problem. 

383. To construct a square equivalent to the sum 
of any number of given squares. 



m- 




Let m, n, o, p, r be sides of the given squares* 

To construct a square :c= m' + ti' + o' +p* + r*. 

Oonstructioiu Take A£ = m. 

Draw AC =n and ± to AB at A, and draw BC. 
Draw CE =0 and ± to BC at C, and draw BE, 
Draw EF =p and ± to BE at E, and draw BF, 
Draw FH=^ r and ± to BF at F, and draw BS, 

The square constructed on BMis the square required. 

Proof. BS^^FW + BT, * 

^FH'+ E0'+EF^ + GT-\-A^, §S79 
(<^ aitm o/ the agttarea on <^ fu>o Uat of art. t^ is equivalent to the square 



on the hypotenuse). 
That is, BH^ «o m* + n* + o* +p' + r». 



Q.B.r. 



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PROBLEMS OF CONSTRUgTION. 196 



Proposition XIV. Problem. 

384. To constmct a polygon similar to two given 
similar polygons and equivalent to their sum. 




1 1 K 



^ 



/ 

J,T ^' p jj 

Let B and JRf be two similar polygons, and AB and 
A!Bf two homologous sides. 

To construct a similar polygon equivalent to B-\- R\ 

OonBtruction. Construct the rt. Z P. 

Take PH=^ A'B\ and PO = AB. 

Draw OJT, and take A*'B" = Off. 
Upon A''B'\ homologous to AB^ construct B" similar to B. 
Then B" is the polygon required. 

Proof. P^ + PS^^OS", .\A^ + A!W=-AjnB^. 

TVT ^ 1^ 
Now -z^. -■= , » 

B" A"B'' 

and ^==4^, §376 

B'* A^f£ff^ 

{rimUar polygons are to each other as the squares of their homologous tides). 
By addition. =K±^'=S±^=1. 



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19G 



PLANE GEOMETRY. — BOOK IV. 



Proposition XV. Problem. 

385. To construct a polygon similar to two given 
similar polygons and equivalent to their difference. 



r- 




< -^ r\ 



/ 



A! B' A U A" 2J" P O 

Let R and R' be two similar polypus, and AB and 
A'B^ two homologous sides. 

To construct a similar polygon equivalent to JR! — R, 
Oonstruction. Construct the rt. Z P, 

and take PO = AB. 
From as a centre, with a radius equal to A^ff, 
describe an arc cutting PX at J?; and join OH. 
Take ^"^" = PH, and on A'*B'\ homologous to AB, 
construct i?" similar to i?. 
Then i?' is the polygon required. 
Prooi PS^=0I£'-0P*, : , AFW'^ ^ .^^ -- aS . 

R' _ AJB'^ 
B" 



Now 



and 



§376 



A^'B^f' 

(similar polygons are to each other as the squares of their homologous tides). 
By subtraction, 



B'-B A^-ABi" 



B" 



A"B''' 



= 1. 



B. 



Q.B.F. 



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PROBLEMS OP CONSTRUCTION. 



197 



/- 



Peoposition XVI. Problem. 

886. To construct a triangle equivalent to a given 
polygon. 




Let ABCDHE be the given polygon. 

To construct a triangle equivalent to the given polygon. 
Oonstraotion. From D draw DE, 

and from JJdraw JIF II to BE. 
Produce A JE to meet B'F at F, and draw DF. 
Again, draw CFy and draw DK II to CF to meet AF pro- 
duced at K, and draw CK. 

In like manner continue to reduce the number of sides of 
the polygon until we obtain the A CIK, 

Proof. The polygon ABCDF has one side less than the 
polygon ABCDHE, but the two are equivalent. 
For the part ABODE la common, 

and the A DEF^ A DEH, § 369 

(Jor the hast DE u common, and their vertices F and H art in the line 
FEW to the base). 

The polygon ABCK has one side less than the polygon 
ABCDFy but the two are equivalent. 

For the part A BCF is common, 
and the A CFK=o^ A CFD, § 369 

{for the base CF is common, and their vertices K and D are in the line 



KD II to the base). 
In like manner the A CIK^ ABCK 



Q.^F. 



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198 PLANE GEOMETRY. — BOOK IV. 



Proposition XVII. Problem. 

887. To construct a square which shall have a given 
ratio to a given square. 




m- 



i / \ 

r 

in ••>, J I 

E"«>. / 

Let R be the given square, and — the given ratio. 

^ m 

To construct a square which shall be to H as n is to m. 

Oonstruction. Take AB equal to a side of B, and draw -4y, 
making any acute angle with AB, 

On Ay take AE=m, EF= n, and join BB. 

Draw FC II to BB to meet AB produced at C, 

On AC OS Q, diameter describe a semicircle. 

At B erect the J. BD, meeting the semi circumference at D, 

Then BI) is a side of the square required. 
Proof. Denote AB by a, BO by b, and BB by x. 

Now a:x = x:b\ that is, x* = ab. § 337 

Hence, a' will have the same ratio to x^ and to ab. 
Therefore a^ : x* = a^ : ab = a : b. 

But a: b = m:n, § 309 

(o straight line drawn through two sides of a A, parallel to the third side, 
divides those sides proportionally). 

Therefore a^ \ a^ = m : n. 

By inversion, a? : a? = n -. m. 

Hence the square on BD will have the same ratio to £ as 
n has to 771. ^^ p. 



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PROBLEMS OF CONSTRUCTION. 199 



Proposition XVIII. Problem. 

388. To construct a polygofn similar to a given poly- 
gon and having a given ra;tio to it. 




\ 

\ 

\ 
\ 
\ 
\ 



Let R be the given polygon and — the given ratio* 

m 

To construct a polygon similar to R, which shall he to R as 
nis to m. 

Oonstruction. Find a line A'B\ such that the square con- 
structed upon it shall be to the square constructed upon A£ 
^ n is to m. § 387 

Upon A*B' as a side homologous to AB, construct the poly- 
gon S similar to R. 

Then S is the polygon required. 

Proof. S:R = A^:AS', §376 

{similar polygons are to each other as the squares of their homologoiis sides). 

But JAg'' : A^ = n:m. Cons. 

# Therefore 8: R — n-.m. 



Ex. 304. Find the area of a right triangle if the length of the hypote- 
nuse is 17 feet, and the length of one leg is 8 feet. 

Ex. 305. Compare the altitudes of two equivalent triangles, if the 
base of one is three times that of the other. 

Ex. 306. The bases of a trapezoid are 8 feet and 10 feet, and the alti- 
tude is 6 feet Find the base of an equivalent rectangle having an equal 
altitude. 



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200 



PLANE aEOMETBY. — BOOK IV. 



Pboposition XIX. Problem. 

889, To construct a square equivalent to a given 
jiarallelogranv. 



R 



U.. 



/ 



\ 



N 



-iyX 



Let ABCD be a parallelogram, b its base, and a its 
altitude. 

To construct a square equivalent to the O ABCD. 

Oonstruotion, Upon the line MX take MN= a, and NO=h. 

Upon MO as a diameter, describe a semicircle. 

At iV erect NP 1. to MO, to meet the circumference at P. 

Then the square R, constructed upon a line equal to NP, 
is equivalent to the O ABCD, 

Proof. MN :NP=NP'. NO, § 337 

(a ± let fall from any point of a circumference to the diameter is a mean 
proportional between the segments of the diameter). 

.\NP'==MNxNO = axb. 

That is, R^OABCD. 

a E. F. 

390. Cor. 1. A square may he constructed equivalent to a 
given triangle, hy taking for its side a mean proportional be- 
tween the base and one-half the altitude of the triangle. 

391. Cor. 2. A squa/re may be constructed equivalent to a 
given polygon, by first reducing the polygon to an equivalent 
triangle, and then constructing a square equivalent to the 
triangle 



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PROBLEMS OF CONSTRUCTION. 201 

Proposition XX. Problem. 

892. To construct a parallelogram equivalent to a 
given square, and, having the sum of its base and 
altitude equal to a given line. 



R 



I 



Xq 



-jn 



Let R be the given square, and let the sum of the 
base and altitude of the required parallelogram be 
equal to the given line MN. 

To construct a O equivalent to R, with the sum of its base 
and altitude equal to MN, 

Oonstruction. Upon MNsiS a diameter, describe a semicircle. 

At -3f erect a ± MF, equal to a side of the given square JR. 

Draw PQ II to MN, cutting the circumference at 8. 

'Dr2iw8C±toMK 

Any O having CM for its altitude and CN for its base is 
equivalent to £, 

Proof. SC= PM. §§ 100, 180 



But 



,\SC* = PM' = P. 
MC:8C=8C'.CN, 



§337 



(a -L hi fall from any 'point in the circumference Co the diameter is a mean 
proportional between the segments of the diameter). 



Then 



SO'^MCxOK 



aE.F. 



Note. This problem may be stated : To construct two straight Hnea 
the sum and product of which are hnovm. 



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202 



PLANE GEOMETBY. — BOOK IV. 



Proposition XXI. Problem. 

393. To construct a parallelogram equivalent to a 
given square, and havin^J the difference of its base 
and altitude equal to a given line. 

s 







V 



Let R be the given square, and let the difference of 
the base and altitude of the required parallelogram 
be equal to the given line MN. 

lb construct a O equivalent to H, with the difference of the 
base and altitude equal to MN. 

Oonstmction. Upon the given line JOT" as a diameter, describe 
a circle. 

From M draw MS, tangent to the O, and equal to a side 
of the given square R. 

Through the centre of the O draw 8B intersecting the cir- 
cumference at Cand B. 

Then any O, as E\ having 8B for its base and 8C for its 
altitude, is equivalent to R. 

Proof. SB:8M=SM:SQ §348 

(if from a point without aOa secant and a tangent are drawn, the tangent is 
a mean jproportional between the whole secant and the part without the O). 

Then SW-SBxSC, 

and the -difference between SB and SO is the diameter of the 
O, that is, JLTiV: q.e.f. 

Note. This problem may be stated : To construct two straight lines 
the difference and product of which are known. 



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PROBLEMS OF CONSTRUCTION. 203 



Proposition XXII. Problem. 

394. To construct a polygon similar to a given poly' 
gon P, and equivalent to a given polygon Q, 



\ ' 1 



I 



m- 



A B" 

Lei P and Q be two polygons, and AB a side of P. 

To construct a polygon similar to P and equivalent to Q. 
Oonstaruction. Find squares equivalent to P and Q, § 391 

and let m and n respectively denote their sides. 
Find A'B\ a fourth proportional to m, n, and AB. § 351 
Upon A^B\ homologous to AB, construct P similar to P. 

Then P' is the polygon required. 
Proof. m'.n=^ AB : A*B\ Cons. 

:.m^,n^^Al^:A!W. 
But P^^m^, and Q=o=w'. Cons. 

... P.Q = m': n^--=AS : ATB^. 

But P',F^AS'.:^E^, §376 

(mnilar polygons are to each other as the squares of their homologous sides). 

.\P'.Q-=--P'.P. Ax. 1 

.*. P is equivalent to Q, and is similar to P by construction. 

as. p. 



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204 



PLANE aEOMETRY. — BOOK IV. 



Problems of Computation. 

Ex. 307. To find the area of an equilateral triangle in terms of its 
. side. 

Denote the side by a, the altitude by h, and the area by 8. 



Then 



But 



A«-a«-^' = ^ 



2 



axh 



g aV3_ fl'\/3 
'2^ 2 " 4 * 




Ex. 308. To find the area of a triangle in terms of its sides. 



By Ex. 219, 
Hence, 



^ =, ^ Vs (8 - a)(« — 6)(s — c). 





b..2. 



flf=|x|V8(8-a)(8-6)(s-c) 
— V« (« — o) (« - b) (« — c). 




Ex. 309. To find the area of a triangle in terms of the radius of the 
circumscribing circle. 

If iJ denote the radius of the circumscribing circle, ^d h the altitude 
of the triangle, we have, by Ex. 222, 0^ 

bxc=2Bxh. 

Multiply by a, and we have 

axbxc = 2Bxaxh. 

But axh '=^28. 

..axbxc=-'AEx8. 

abc 



.'.8= 



4i? 




E< -^ 



Non. The radius of the circumscribing circle is equal to --^ 



abc 



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EXEBCISES. 205 



Theorems. 



\ 



310. In a right triangle the product of the legs is eqnal to the product 
of the hypotenuse and the perpendicular drawn to the hypotenuse from 
the vertex of the right angle. 

311. If ABO ia a right triangle, the vertex of the right angle, 
BD a line cutting AC in D, then BD^ + AQ*^AB^ + VO^. 

312. Upon the sides of a right triangle as homologous sides three 
similar polygons are constructed. Prove that the polygon upon the 
hypotenuse is equivalent to the sum of the polygons upon the legs. 

313. Two isosceles triangles are equivalent if their legs are equal each 
to each, and the altitude of one is equal to half the base of the other. 

yX 314. The area of a circumscribed polygon is equal to half the product 
of its peslmeter by the radius of the inscribed circle. 

315. Two parallelograms are equ^iL^jA adjacent sides of the one 
are equal respectively to two adjacent 8i4pf the other, and the included 
angles are supplementary. '^'^^^^IM^ 

316. Every straight line drawn through the centre of a parallelogram 
divides ^ into two equal parts. 

317/ If the middle points of two adjacent sides of a parallelogram are 
joined, a triangle is formed which in equivalent to one-eighth of the 
entire parallelogram. 

318. If any point within a parallelogram is joined to the four vertices, 
the sum of either pair of triangles having parallel bases is equivalent to 
one-half the parallelogram. 

319. The line which joins the middle points of the bases of a trap*- 
aoid divides the trapezoid into two equivalent parts. 

J> 320. The area of a trapezoid is equal to the product of one of the legs 
and the distance from this leg to the middle point of the other leg. 

321. The lines joining the middle point of the diagonal of a quadri- 
lateral to the opposite vertices divide the quadrilateral into two equiva- 
lent parts. 

\^ 322. The figure whose vertices are the middle points of the sides of 
any quadrilateral is equivalent to one-half of the quadrilateral. 
\ 323. ABO la a triangle, M the middle point of AB, P any point in 
AB between A and M. If MD is drawn parallel to PC, and meeting 
BOiX D, the triangle BFD is equivalent to one-half the triangle ABG. 



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206 * PLANE GEOMETRY. — BOOK IV. 



NuHEBiGAL Exercises. 

324. Find the area of a rhombus, if the sum of its diagonals is 12 feet, 
and their ratio is 3 ; 5. g ^ ^19 

325. Find the area of an isosceles right triangle if the hypotenuse 
is 20 feet U 

326. In a right triangle, the hypotenuse is 13 feet, one leg is 5 feet 

Find the area. < ^ , -^^r^ 

327. Find the area of an isosceles triangle if the base — 6, and leg » c. . tdt * 

328. Find the area of an equilateral triangle if one side "• 8. r \# ^\ ' . 

329. Find the area of an equilateral triangle if the altitude = h. u>r^ 

330. A house is 40 feet long, 30 feet wide, 25 feet high to the eaves, ^ 
and 35 feet high to the ridge-pole. Find the number of square^et in ^ ^^ ^ 
its entire exterior surface. '^^ 



[uare^i 

331. The sides of a right triangle are as 3 : 4 : 5. The alti^Te upon 
the hypotenuse is 12 feet. Find the area. 

332. Find the area of a right triande if one leg »• a, and the altitude 
upon the hypotenuse = A. ^ THTt— 

333. Find the area of a triangle ii the lengths of the sides are 104 
feet, 111 feet, and 175 feet • / ^ 

^ 334. The. area of a trapezoid is "/OO square feet The bases are 30 feet 
and 40 feet respectively. Find the distance between the bases. 

335. ABCD is a trapezium; ^^ = 87 feet, ^(7= 119 feet, CD = 41 
feet, DA - 169 feet, ilC=- 200 feet Find the area. 

336. What is the area of a quadrilateral circumscribed about a circle 
whose radius is 25 feet, if the perime^^er of the quadrilateral is 400 feet ? 
What is the area of a hexagon havings an equal perimeter and circum- 
scribed about the same circle ? . ' *' 

337. The base of a triangle is 15 feet, and its altitude is 8 feet Find 
the perimeter of an equivalent rhombus if the altitude is 6 feet. / 

338. Upon the diagonal of a rectangle 24 feet by 10 feet a triangle 
equivalent to the rectangle is constructed. What is its altitude ? 

339. Find the side of a square equivalent to a trapezoid whose bases 
are 56 feet and 44 feet, and each leg is 10 feet I y • ) 

340. Through a point Pin the side AB of a triangle ABC, a line is 
drawn parallel to BC, and so as to divide the triangle into two equiva- 
lent parts. Find the value of AP in terms of AB. 



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EXERCISES. 207 

341. What part of a parallelogram is the triangle cut off by a line 
drawn from one vertex to the middle point of one of the opposite sides ? 

342. In two similar polygons, two homologous sides are 15 feet and 
25 feet. The area of the first polygon is 450 square feet. Find the area 
of the other polygon. 

343. The base of a triangle is 32 feet, its altitude 20 feet. What is 
the area of the triangle cut off by drawing a line parallel to the base 
and at a distance of 15 feet from the base ? 

344. The sides of two equilateral triangles are 3 feet and 4 feet. Find 
the side of an equilateral triangle equivalent to their sum. 

345. If the side of one equilateral triangle is equal to the altitude of 
another, what is the ratio of their areas ? 

346. The sides of a triangle are 10 feet, 17 feet, and 21 feet. Find 
the akfi of the parts^into which the triangle is divided by bisecting the 
angle^P^ed by the first two sides. 

347. In a trapipoid, one base is 10 feet, the altitude is 4 feet, the area 
18 32 square feet^ Find the length of a line drawn between the legs 
parallel to the base and distant 1 foot from it. 

348. If the altitude A of a triangle is increased by a length m, how 
much must be taken from the base a in order that the area may remain 
the same ? 

349. Find the area of a right triangle, having given the segments p, 
q, into which the hypo teniae is divided by a perpendicular drawn to the 
hypotenuse from the vertex ..of the right angle. 

Problems. 

850. To construct a triangle equivalent to a given triangle, and 
having one side equal to a given. length I. 

351. To transform a triangle into an equivalent right triangle. 

352. To transform a triangle into an equivalent isosceles triangle. 

353. To transform a triangle ABO into an equivalent triangle, hav- 
ing one side equal to a given length I, and one angle equal to angle BAC, 

Hints. Upon AB (produced if necessary), take AD = I, draw BE II to 
CD, and meeting -4 C (produced if necessary) at ^; A BEDoABEO. 

354. To tran^orm a given triangle into an equivalent right triangle, 
having one leg equal to a given length. 



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208 PLANE GEOMETRY. — BOOK IV. 

856. To transform a given triangle into an equivalent right triangle, 
having the hypotenuse equal to a given length. 

356. To transform a given triangle into an equivalent isosceles tri- 
angle, having the base equal to a given length. 

' ^ construct a triangle equivalent to : 

357. The sum of two given triangles. 

358. The difference of two given triangles. 

359. To transform a given triangle into an equivalent equilateral 
triangle. 

To transfonn a parallelogram into : 

360. A parallelogram having one side equal to a given length. 

361. A parallebgram having one angle equal to a given angle. 

362. A rectangle having a given altitude. 

To transform a square into : mM 

363. An equilateral triangle. ^ 

364. A right triangle having one leg equal to a given length. 

365. A rectangle having one side equal to a given length. 

To construct a square equivalent to : 
^^66. Five-eighths of a given square. 
>867. Three-fifths of a given pentagon. 
^368. To draw a line through the vertex of a given triangle so. as to 
divide the triangle into two triangles which shall be to each other as 2 : 3. 
f 369. To divide a given triangle into two equivalent parts by drawing 
a line through a given point P in one of the sides. 

370. To find a point within a triangle, such that the lines joining this 
point to the vertices shall divide the triangle into three equivalent parts, 
^^71. To divide a given triangle into two equivalent parts by drawing 
a line parallel to one of the sides. 

372. To divide a given triangle into two equivalent parts by drawing 
a line perpendicular to one of the sides. 

>373. To divide a given parallelogram into two equivalent parts by 
drawing a line through a given point in one of the sides. 

374. To divide a given trapezoid into two equivalent parts by draw- 
ing a line parallel to the bases. 

^75. To divide a given trapezoid into two equivalent parts by draw- 
ing a line through a given point in one of the bases. 



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BOOK V. 

REGULAR POLYGONS AND CIRCLES. 

895. A regular polygon is a polygon which is equilateral 
and equiangular ; as, for example, the equilateral triangle, and 
the square. 

Proposition I. Theorem. 

396. An equilateral polygon inscribed in a circle is 
a regular polygon. 




Let ABC etc., be an equilateral polygon inscribed in 
a circle. 

To prove the polygon ABC etc., regular. 

Proof. The arcs AB, BC, CD, etc., are equal, § 230 

{in the same O, equal chords subtend equal arcs). 

Hence arcs ABC, BCD, etc., are equal. Ax. 6 

and the A A, B, C, etc., are equal, 

(being inscribed in equal segments). 

Therefore the polygon ABC, etc., is a regular polygon, being 
equilateral and equiangular. a t o. 



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210 PLANE GEOMETRY. — BOOK V. 



Pboposition II. Theorem. 

397. A circle may be circumscribed about, and a 
circle m^y be inscribed in, any regular polygon. 







.__F 

Let A^BGDE he a regular polygon. 
I. To prove that a circle may he circumscribed aboict 
ABODE, 
Proof. Let be the centre of the circle passing through 

A, B, a 

Join OA, OB, 00, and OD. 

Since the polygon is equiangular, and the A OBOis isosceles, - 

ZABO=-ZBOI), 
and ZOBO = ZOOB. § 154 

By subtraction, Z OB A = Z OOD. 

Hence in the A OB A and 001) 

the ZOBA = Z OOD, 

the radius OB = the radius 00, 

and AB=OD. §395 

.\AOAB = AOOD, §150 

{having two sides and the included A of the one equal to two sides and the 
included /.of the other). 

/. OA = OD. 

Therefore the circle passing through A, B, and O, also 

passes through D. 



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REGULAR POLYGONS AND CIRCLES. 211 

In like manner it may be proved that the circle passing 
through B^ (7, and i>, also passes through E\ and so on 
through all the vertices in succession. 

Therefore a circle described from as a centre, and with a 
radius OAj will be circumscribed about the polygon. 

II. To prove that a circle may he inscribed in ABCDE, 

Proof. Since the sides of the regular polygon are equal 
chords of the circumscribed circle, they are equally distant 
from the centre. - § 236 

Therefore a circle described from as a centre, and with 
the distance from to a side of the polygon as a radius, will 
be inscribed in the polygon. aE.D. 

898. The radius of the circumscribed circle, OA, is called 
the radius of the polygon. 

899i The radius of the inscribed circle, 0-P, is called the 
apothem of the polygon. 

400. The common centre of the circumscribed and in- 
scribed circles is called the centre of the polygon. 

401. The angle between radii drawn to the extremities of 
any side, as angle AOB^ is called the angle at the centre of the 
polygon. 

By joining the centre to the vertices of a regular polygon, 
the polygon can be decomposed into as many equal isosceles 
triangles as it has sides. Therefore, 

402. Cor. 1. The angle at the centre of a regular polygon is 
eqvAxl to four tight angles divided hy the number of sides of 
the polygon. 

403. Cor. 2. The radius drawn to any vertex of a regular 
polygon bisects the angle at the vertex. 

404. Cor. 3. The interior angle of a regular polygon is the 
supplement of the angle at the centre. 

For the A ABO =2 Z ABO = Z ABO + ZBAO. Hence 
the Z ABO IB the supplement of the Z AOB, 



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212 PLANE GEOMETRY. — BOOK V. 

Proposition III. Theorem. 

405. If the circumference of a circle is divided into 
any number of equal parts, the chords joining the 
successive points of division form a regular inscribed 
polygon, and the tangents drawn al the points of 
division form a regular circumscribed polygon. 

I D H 




F 

Let the circumference be divided into equal arcs, 
AB, BC, CD, etc., be chords, FBQ, GCH, etc., be tangents. 

I. To prove that ABODE is a regular polygon. 

Proof. The sides AB, BC, CD, etc., are equal, § 230 

{in the same O egtuil arcs are subtended by equal chords). 

Therefore the polygon is regular, § 896 

{an equilateral polygon inscribed in a O is regular). 

II. To prove that the polygon FOHIKis a regular polygon. 
Proof. In the A AFB, BOO, CUD, etc. 

AB=BO=: CD, etc. §896 

Also, ZBAF=:ZABF=ZCBQ = ZBCO, etc., §269 
{being measured by halves of equal arcs). 

Therefore the triangles are all equal isosceles triangles. 

Hence ZF^AO^AH, etc. 

Also, FB^^BO^OC = CH, etc. 

Therefore FO = OB', etc. 

.'. FOHIK'is a regular polygon. § 895 

Q.B.a 
406. Cor. 1. Tangents to a circumference cU the vertices of a 
regular inscnbed polygon form a regular circumscribed poly- 
gon of the same number of sides. 



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REGULAR POLYGONS AND CIRCLES. 



213 




407. Cor. % If a regular polygon is inscribed in a circle, 
the tangents drawn at the middle points 
of the arcs sitbtended by the sides of the 
polygon form a circumscribed regular 
polygon, whose sides are parallel to the 
sides of the inscribed polygon and whose 
vertices lie on the radii (prolonged) of 
the insciibed polygon. For any two cor- J? m b' 
responding sides, as AB and A^B\ perpendicular to OM, 
are parallel, and the tangents MB^ and NB\ intersecting at a 
point equidistant from OJf and OiV(§ 246), intersect upon the 
bisector of the Z MON{% 163) ; that is, upon the radius OB, 

408. CoR. 3. If the vertices of a regular inscribed polygon 
are joined to the middle points of the arcs sub- 
tended by the sides of the polygon, the joining 
lines form a regular inscribed polygon of 
double the number of sides. 

409. Ck)R. 4. If tangents are drawn at the 
middle points of the arcs between adjacent, 
points of contact of the sides of a regular cir- 
cumscribed polygon, a regular circumscribed 
polygon of double the number of sides is 
formed, 

410. Scholium. The perimeter of an inscribed polygon is 
less than the perimeter of the inscribed polygon of double the 
number of sides; for each pair of sides of the second polygon 
is greater than the side of the first polygon which they replace 
(§ 137). 

The perimeter of a circumscribed polygon is greater than 
the perimeter of the circumscribed polygon of double the num- 
ber of sides ; for every alternate side FO, HI, etc., of the poly- 
gon FOHI, etc., replaces portions of two sides of the circum- 
scribed polygon ABCD, and forms with them a triangle, and 
one side of a triangle is less than the sum of the other two sides. 




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214 



PLANE GEOMETRY. — BOOK V. 



Proposition IV. Theorem. 

411. Two regular polygcns of the same number of 
sides are similar. 





A B A' 

Let Q and 01 be two regular polygons, each having 
n sides. 
To p'ove Q and Q similar polygons. 

Proof. The sum of the interior A of each polygon is equal to 

(n-2)2rt..^, §205 

ifhe mm of the interior A of a polygon u equal to 2 rt A taken ob many 
times less 2 as the polygon has sides). 

Each angle of either polygon = \ ~~ } — : — , § 206 

n 
(for the A of a regular polygon are all equal, and hence each Z is equal 
to the sum of the A divided By their number). 

Hence the two polygons Q and Q* are mutually equiangular. 

Since A£ = £C, etc., and A'B' = B'0\ etc., § 895 

ABiA'B'^^BOiB'C^eUi. 

Hence the two polygons have their homologous sides 



proportional. 
Therefore the two polygons are similar. 



§319 
aE.0. 



412. Cor. The areas of two regular polygons of the same 
number of sides are to each other as the squares of any two 
homohgovs sides, § 376 



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REGULAR POLYGONS AND CIRCLES. 



215 



Proposition V. Theorem. 

418. The perimeters of two regular polygons of the 
same number of sides are to ea£h other as the radii 
of their eircumscribed eircles, and also as the radii 
of their inscribed circles. 

D 





A M B A' M' B' 

Let P and P' denote the perimeters, and (/ the 
centres, of the two regular polygons. 

From 0, 0' draw 0-4, (7^', 05, (y£\ and Ja OM, aMK 
To prove F:P = OA: 0'^' = OM: O'ML 
Proof. Since the polygons are similar, 

F:F' = AB:A'B\ 
In the isosceles A GAB and OA^B^ 

the Z = the Z 0', 
andO^:05=0'^':0'5'. 
/. the A GAB and OA^B^ are similar. 
.\AB'.A'B'=GA:aA\ 

Also AB : A'B^ = GM: aM\ 

(the homologous altitudes of similar ^ have the same ratio as their bases). 
.-. F:F>=OA: a A' = OM: OM'. 

Q.E.O. 

414 OoE. The areas of two regular polygons of the same 
number of sides are to each other as the squares of the radii 
of their circumscribed circles, and also as the squares of the 
radii of their inscribe circles. § 376 



§411 
§883 

§402 

§326 
§319 
§328 



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216 



PLANE GEOMETRY. — BOOK V. 



Proposition VI. Theorem. 

415. Th^ difference between the lengths of the peHinn 
eters of a regular inscribed polygon and of a similar 
circumscribed polygon is indefinitely diminished as 
the nwmber of the sides of the polygons is indefinitely 
increased. 




Let P and P' denote the lengths of the perimeters, 
AB and A'B* two corresponding sides, OA and OA' the 
radii, of the polygons. 

To prove that as the number of the sides of the polygons is 
indefinitely increased^ P — P is indefinitely diminished, 

Proof. Since the polygons are similar, 

P'.P=OA':OA. §333 

Therefore P — P : P :: OA' — OA: OA. § 301 

Whence OA{P — P) = P{pA' — OA). § 295 

Now OA is the radius of the circle, and P, though an 

increasing variable, always remains less than the circumference 

of the circle. 

Therefore P-^ Pis indefinitely diminished, if OA^ — OA 

is indefinitely diminished. 

Draw the radius 00 to the point of contact of A'B', 
In the A OA'C, OA' - 00 < A'C. ' § 137 

Substituting OA for its equal 0(7, we have 
OA'-OA<A'C. 



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EEGULAB POLYGONS AND CIRCLES. 217 

But as the number of sides of the polygon is indefinitely 
increased, the length of each side is indefinitely diminished ; 
that is, A^B\ and consequently A^C, is indefinitely diminished. 

Therefore OA^ — • OA, which is less than A^C, is indefinitely 
diminished. 

Therefore P'— P is indefinitely diminished. a e. d. 

416. Cor. The difference between the areas of a regular 
inscribed polygon and of a similar circumscribed polygon is 
indefinitely diminished as the number of the sides of the poly- 
gons is indefinitely increased. 

For, if 8 and /S" denote the areas of the polygons, 

8' -.8= OAf : OT = OZ'' : OCT. § 414 

By division, 8^-8:8= OA^ - 6o' : 0C\ 



Whence 8' -8= 8x ^'^' Z.?^ =8x^'^ 



Since A^O can be indefinitely diminished by increasing the 
number of the sides, 8 — 8 can be indefinitely diminished. 

417. Scholium. I'he perimeter P' is constantly greater 
than P, and the area 8^ is constantly greater than 8\ for the 
radius OA' is constantly greater than OA. But P' constantly 
decreases and P constantly increases (§ 410), and the area 8' 
constantly decreases, and the area 8 constantly increases, as 
the number of sides of the polygons is indefinitely increased. 

Since the difierence between P' and P can be made as 
small as we please, but cannot be made absolutely zero, and 
since P' is decreasing while P is increasing, it is evident that 
P' and P tend towards a common limit. This common limit 
is the length of the circumference. § 259 

Also, since the difierence between the areas 8* and 8 can be 
made as small as we please, but cannot be made absolutely 
zero, and since /S" is decreasing, while 8 is increasing, it is 
evident that /S" and 8 tend towards a common limit. This 
common limit is the area of the circle. 



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218 



PLANE GEOMETRY. — BOOK V. 



Proposition VII. Theorem. 

418. Two circumferences have the same ratio as 
their radii. 





Let C and C be the circumferences, R and S' the 
radii, of the two circles Q and Of. 

To prove G\C^=R'.RK 

Proof. Inscribe in the ® two similar regular polygons, and 
denote their perimeters by P and P, 

Then P:P = P: i2'(§ 413); that is, ^ X P=^X P*. 

Conceive the number of the sides of these similar regular 
polygons to be indefinitely increased, the polygons continuing 
to have an equal number of sides. 

Then Rx P will continue equal to iZx P, and P and P 
will approach indefinitely O and C as their respective limits. 

.-. P'X(7=-PXC'(§260); that is, C: C' = P:B'. 

aca 

419. Cor. 27ie ratio of the circumference of a circle to its 
diameter is constant. For, in the above proportion, by doubling 
both terms of the ratio P : P\ we have 

0:0' =2P:2P'. 
By alternation, 0:2 P=0' :2RK 
This constant ratio is denoted by tt, so that for any circle 
whose diameter is 2 P and circumference C, we have 
J^ 
2R 

420. Scholium. The ratio tr is incommensurable, and there- 
fore can be expressed in figures only approximately. 



: = 7r, or (7=2irP. 



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REGULAR POLYGONS AND CIRCLES. 219 

Proposition VIII. Theorem. 

421. The area of a regular polygon is equal to one' 
half the product of its apothem by its perimeter. 

E 





A U B 

Let P represent the perimeter, R the apothem, and 
8 the area of the regular polygon ABC etc. 

To prove 8=\RxP. 

Proof. Draw OA, OB, 0(7, etc. 

The polygon is divided into as many A as it has sides. 

The apothem is the common altitude of these A, 

and the area of each A is equal io ^ E multiplied by the 
base. § 368 

Hence the area of all the A is equal to ^ ^ multiplied by 
the sum of all the bases. 

But the sum of the areas of all the A is equal to the area 
of the polygon. 

and the sum of all the bases of the A is equal to the perim- 
eter of the polygon. 

Therefore /S=4^X P. 

aE.D. 

422. In different circles similar arcs, similar sectors, and 
similar segments are such as correspond to equal angles at 
the centre. 



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220 PLANE GEOMETRY. — BOOK V. 



Proposition IX. Theorem. 

423. The area of a circle is equal to one^half the 
product of its radius by its circumference. 

E 




BMC 
Let B represent the radius, the circumference, 
Bnd S the area, of the circle. 

To prove 8=^RxC, 

Proof. Circumscribe any regular polygon about the circle, 
and denote its perimeter by P. 

Then the area of this polygon = | P X P, § 421 

Conceive the number of sides of the polygon to be indefi- 
nitely increased ; then the perimeter of the polygon approaches 
the circumference of the circle as its limit, and the area of the 
polygon approaches the circle as its limit. 

But the area of the polygon continues to be equal to one- 
half the product of the radius by the perimeter, however great 
the number of sides of the polygon. / 

Therefore 8^\E xd, § 260 

/ aE.D. 

424. Cor. 1. The area of a sector equals one-half the product 
of its radius hy its arc. For the sector is such a part of the 
circle as its arc is of the circumference. 

425. Cor. 2. The area of a circle equals w times the square 
of its radius. 

For the area of the O ^ ^ P x (7=^^ X 2irB = ir£^. 



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REGULAR POLYGONS AND CIRCLES. 221 

426. CoR. 3. The areas, of two circles are to each othe^' as the 
squares of their radii. For, if S and S* denote the areas, and 
H and H' the radii, 

427. Cor. 4. Similar arcs, being like parts of their respective 
circumferences, are to each other as their radii; similar secfxyrs, 
being like parts of their respective circles, are to each other as 
the squares of their radii. 

Proposition X. Theorem. 

428. The areas of two similar segments are to ea^h 
other as the squares of their radiL 

C 



Let AC andA'C be the radii of the two similar seg- 
ments ABP and A'B'Ff. 

To prove ABP : A'B'P = AQ^ : A^. 

Proof. The sectors ACB and A*Q^B^ are similar, § 422 
{having the A at the centre, Qand (7, equaJ). 

In the A ACB and A'C*B^ 

AQ=A C\ AQ= CB, and A'O'^C'B'. 

Therefore the A ACB and A'C^B* are similar. § 326 

Now sector A CB : sector A'C^B' = AC^ : A^C^\ § 427 

and A ACB: A A'C'B' = AC' : jJC^, § 375 

Hence sector ^Og^ A ^^^ ^Zg, g 3^^ 

sector ^'C''^' - A ^'C'^' JTqi^ ^ 



That is, ABF : A'B'F^ = AG' : AJG^. 



Q.E.a 



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222 PLANE GEOMETRY. — BOOK V. 

Problems of Construction. 
Proposition XI. Problem. 
429. To inscribe a square in a given circle. 



Let be the centre of the given circle. 

To inscribe a square in the circle, 

Oonstruction. Draw the two diameters AC and BD ± to 
each other. 

Join AB, BC, CD, and DA. 

Then ABCD is the square required. 

Proof. The A ABC, BCD, etc., are rt. A, § 264 

{Jbeing inscribed in a semicircle), 

and the sides AB, BC, etc., are equal, § 280 

{in the same O eqiLal arcs are subtended by equal chords). 

Hence the figure ABCD is a square. § 171 

aE.F. 

430. Cor. By bisecting the arcs AB, BC, etc., a regular 
polygon of eight sides may be inscribed in the circle ; and, by 
continuing the process, regular polygons of sixteen, thirty-two, 
sixty-four, etc,, sides may be inscribed. 



Ex. 376. The area of a circumscribed square is equal to twice the 
area of the inscribed square. 

Ex. 377. If the length of the side of an inscribed square is 2 inches, 
what is the length of the circumscribed square ? 



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PROBLEMS OP CONSTBUCTION. 223 

Proposition XII. Problem. 
431. To inscribe a regular hexagon in a given circle. 



Let be the centre of the given circle. 

To insc9;ihe in the given circle a regular hexagon. 
OonstniotioiL From draw any radius, as 00. 

From (7 as a centre, with a radius equal to OC, 
describe an arc intersecting the circumference at K 

Draw Oi^ and OF. 
Then OF is a side of the regular hexagon required. 
Proof. The A OFO is equilateral and equiangular. 
Hence the Z FOOia ^ of 2 rt. A, or ^ of<4 rt. A. § 138 
And the arc FOis ^ of the circumference ABOF. 
Therefore the chord FO, which subtends the arc FO, is a 
side of a regular hexagon ; 

and the figure OFD etc., formed by applying the radius six 
times as a chord, is a regular hexagon. o. e. f. 

432. Cor. 1. By joining the aUemate vertices A, O, D, an 
equilat€7*al triangle is inscribed in the circle. 

433. Cor. 2. By bisecting the arcs AB, BO, etc., a regular 
polygon of twelve sides may be inscribed in the circU ; and, by 
continuing the process, regular polygons of twenty-four, forty- 
eight, etc., sides may be inscribed. 



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224 PLANE aEOMETRY. — BOOK V. 

Proposition XIIL Problem. 
484. To inscribe a regular decagon in a given circle. 




B 
Let he the centre of the given circle. 

To insci'ihe a regular decagon in the given circle, 

OonstniotioiL Draw the radius OC, 

and divide it in extreme and mean ratio, so that 00 shall 
be to O/S.as 08 is to 80. § 356 

From (7 as a centre, with a radius equal to 08, 

describe an arc intersecting the circumference at B, and 
draw BO. 

Then BO is a side of the regular decagon required. 

Proof. Draw B8 and BO. 

Bj construction 00:08=08:80, 
and BO =08. 

.-. 00:B0=B0:8a 
Moreover, the Z OOB = Z 8CB. Iden. 

Hence the A OOB and B08 are similar, . § 326 

(having an Z of the one equal to an Z of the other, and the including sides 
proportional). 

But the A OOB is isosceles, 

(its sides 00 and OB being radii of the same circle). 

/. A B08, which is similar to the A OOB, is isosceles, 

and 0B = B8=08. 



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PROBLEMS OF CONSTRUCTION. 225 

.-. the A SOB is isosceles, and the Z = Z SBO. 

Buttheext.ZGS^ = Z0 + Z/?50 = 2Z0. §145 

Hence Z SOB (= Z CSB) = 2 Z O, § 164 

and Z OBC(= Z /S(7^) = 2 Z O, § 154 

/. the sum of the A of the A OCB = bZ = 2rt. A, 

and Z = I of 2 rt. ^, or ^ of 4 rt. A. 

Therefore the arc BC ia -^ oi the circumference, 

and the chord BCisB. side of a regular inscribed decagon. 

Hence, to inscribe a regular decagon, divide the radius in 
extreme and mean ratio, and apply the greater segment ten 
times as a chord. 

Q.E.F. 

436. Cor. 1. Bi/ joining the altemcde vertices of a regular 
inscribed decagon^ a regular pentagon is inscribed. 

436. Cor. 2. By bisecting the arcs BQ, CF, etc., a regular 
polygon of twenty sides may he inscribed; and, by continuing 
the process, regular polygons of forty, eighty, etc., sides may be 



Let R denote the radius of a regular inscribed polygon, r the apothem, 
a one side, A an interior angle, and Q the angle at the centre ; show that 

Ex. 378. In a regular inscribed triangle a = R V3, r => J iJ, -4 =» 60°, 
<7- 120*>. 

Ex.379. In an inscribed square a = RV2, r = \R^/2, il=-90*», 
C-90O. 

Ex. 380. In a regular inscribed hexagon a'=^R,r='i^R >/3, A = 120**, 
0-60°. 

Ex. 381. In a regular inscribed decagon 

a«-5L-Izi}, r = Ji2Viq + 2\/5. A =144°. C-36°. 



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226 PLANE GEOMETRY. — BOOK V. 



Peoposition XIV. Peoblem. 

487. To inscribe in a given circle a regular penis' 
decagon, or polygon of fifteen sides. 



F 

Let Q be the given circle. 

To inscribe in Qa regular pentedecagon, 

ChmstractioiL Draw ^jET equal to a side of a regular inscribed 
hexagon, § 431 

and ^^ equal to a side of a regular inscribed decagon. § 434 

Join FH, 

Then -MT will be a side of a regular inscribed pentedecagon. 

Proof. The arc EH is \ of the circumference^ 

and the arc EF is -^ of the circumference. 

Hence the arc FH is |- — -j^, or ^, of the circumference, 

and the chord FH is a side of a regular inscribed pente- 
decagon. 

By applying FH fifteen times as a chord, we have the 
polygon required. 

438. Cob. By bisecting the arcs FH, HA, etc., a regular 
polygon of thirty sides may be inscribed; and, by cordinuing 
the process, regular polygons of sixty, one hundred twenty, etc., 
sides, may be inscribed. 



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PEOBLEMS OF CONSTRUCTION. 227 



Peoposition XV. Peoblem. 

439. To inscribe in a given circle a regular polygon 
similar to a given regular polygon. 




Let ABCD etc., he the given regular polygon, and 
CD'S' the given circle. 

To inscribe in the circle a regular polygon similar to ABCD, 
etc. 

Oonstniction. From 0, the centre of the given polygon, 

draw on and 00. 

From 0', the centre of the given circle, 

draw aC and ffD', 

making the Z 0^ = Z0. 

Draw a'i)'. 

Then C^D* will be a side of the regu 

Proof. Each polygon will have as 
(= Z 0') is contained times in 4 rt. A. 

Therefore the polygon C'D'U' etc., 
gon CBU etc., 

(two regtUar polygons of the same number of sides are similar). 

Q.E. p. 



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228 PLANE GEOMBTBY. — BOOK V. 



Peoposition XVI. Problem. 

440. Given the radius and the side of a regular 
inscribed polygon, to find the side of the regular 
inscribed polygon of double the number of sides. 

D 




LetAB be a side of the regular inscribed polygon. 

To find the value of AD, a side of a regular inscribed poly- 
gon of dovhle the number of sides. 

From D draw Z^JT through the centre O, and draw OA, AH. 
DHi8 ± to A£ at its middle point C. § 123 

In the rt. A OA C, 6C^= OJl - ACf, § 839 

That is, 00=^01^-10^. 

«AO=^\AB\ henceZc'^jZs'. 
re, 00=y/6A^-\AS. 

•t. A BAH, § 264 

AD^^DHxDO §334 

^-^OAipA-OO), 
and AB = V20^(0^- 0(7). 

If we denote the radius by R, and substitute ViJ^ — iXS* 
for 00, then 



= A/^(2^-V4i?-ZS^. 



Q.E.P. 



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PEOBLEMS OP COMPUTATION. 



A 



229 



Peoposition XVII. Peoblem. 

441. To compute the ratio of the cirewmference of a 
circle to its diameter approximateLy* 




Let C $e the circumference, and B the radius. 

Tofincfihe numerical value of v. 

2,ri2=a §419 

Therefore when jS = 1, ir = \C. 
"We make the following computations by the use of the 
formula obtained in the last proposition, when -B = 1, and 
AB = 1 (a side of a regular hexagon). 

No. 
BIdM. Form of Compntatlon. Length of Side. Length of Perimeter. 

12 Cx=V2-V4^« 0.51^6^09, 6.21165708 

24 c,= V2- V4 - (0.517638097 0.26105238 M||^5722 

S396 



48 <?,= V2~ V4 - (0.26105238/ 0.13080626 




96 C4= V2- V4- (0.13080626)* 0.06543817 

192 ^6= V2- V4-(0.06543817y 0.03272346 6.28290510 

384 cg= V2- V4-(0.032723467 0.01636228 6.28311544 

768 c,= V2- V4- (0.01636228^ 0.00818121 6.28316941 

Hence we may consider 6.28317 as approximately the cir- 
cumference of a O whose radius is unity. 

Therefore ir = ^(6.28317) = 3.14159 nearly. q.e.p. 
442. Scholium. In practice, we generally take 
«• = 3.1416, i = 0.31831. 

IT 




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230 



PLANE GEOMETRY. — BOOK V. 



Maxima and Minima. — Supplement abt. 

443. Among magnitudes of the same kind, that which is 
greatest is the maximum^ and that which is smallest is the 
minimum. 

Thus the diameter of a circle is the maximum among all 
inscribed straight lines ; and a perpendicular is the minimum 
among all straight lines drawn from a point to a given line. 

444. Isoperimetric figures are figures which have equal 
perimeters. 

Peoposition XVIII. Theorem. 

445. Of all triangles having two given sides, that 
in which these sides include a right angle is the 
Tuaximum, 



^N 


E 




iV 




K„ 


D B 


I) 


^u 



Let the triangles ABC and EBC have the sides AB 
and BC equal respectively to EB and BC ; and let the 
angle ABC be a right angle, 

lb prove A ABC > A UBO. 

Proof. From E let fall the ± ED, 

The A ABO and EBC, having the same base BC, are to 

each other as their altitudes AB and ED, § 370 

Now EB > ED. § 114 

By hypothesis, EB = AB. 

.'. AB > ED. 

.\AABC>AEBO. ^^q 



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MAXIMA AND MINIMA. 



231 



Proposition XTX. Theorem. 

446. Of all triangles haxdng the same hose and equal 
perimeters, the isosceles triangle is the m^oMimMm,. 




Let the AACB and ADB have equal perimeters, and 
let the A ACB be isosceles. 

To prove A ACB > A ADB. 

Proof. Produce AC to IT, making 0^= AC, and join SB. 
ABH is a right angle, for it will be inscribed in the semi- 
circle whose centre is (7, and radius CA. 

Produce SB, and take DP= DB. 

Draw O^and DM II to AB, and join AP. 

Now AH= AC+ GB=^AD+DB=AD + DP. 

B\iiAD + DP>AP, hence AH> AP. 

Therefore HB > BP. § 120 

But KB = iSB and MB=:iBP. § 121 

Hence KB > MB. 
By § 180, KB=CI! and MB=DF, the altitudes of the 
A ACB and ADB. 

Therefore A ABC> A ADB. § 370 



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232 PLANE aEOMETEY. — BOOK V. 



Pboposition XX. Theorem. 

447. Of all pcHygcns wUh sides allMven hut one, 
the maximum can be inscribed in a semicircle which 
has the undetermined side for its diamster. 

o \ 




Let ABCDE be the marimnm of polygons with sides 
AB, BC, CD, DE, and the extrezaities A and E on the 
straight line MN. 

To prove ABCDE can he inscribed in a semicircle. 

Proof. From any vertex, as (7, draw CA and CE, 
The A ACE must be the maximum of all A having the 
given sides CA and CE\ otherwise, by increasing or diminish- 
ing the Z, ACE, keeping the sides CA and CE unchanged, but 
sliding the extremities A and E along the line MN, we can 
increase the A ACE, while the rest of the polygon will remain 
' unchanged, and therefore increase the polygon. 

But this is contrary to the hypothesis that the polygon is 
the maximum polygon. 

Hence the A ACEvtith. the given sides CA and CE is the 
maximum. 

Therefore the Z ACE is a right angle, § 445 

(the maximum of ^ having two given sides is the A with the two given sides 
including a rU Z). 

Therefore Clies on the semi-circumference. § 264 

Hence every vertex lies on the circumference ; that is, the 

maximum polygon can be inscribed in a semicircle having the 

undetermined side for a diameter. « q.e.o. 



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MAXIMA AND MINIMA, 



233 



Pboposition XXI. Theorem. 

448. Of all polygcfns with given sides, tfuvt which 
can he inscribed in a circle is the maximum. 





Let ABODE be a polygon inscribed in a circle, smd 
A'S^C^D^E' be a polygon, equilateral with respect to 
ABODE, which cannot be inscribed in a circle. 

To prove' ABODE greai&r than A^B*C*I)^E\ 

Proof. Draw the diameter AH, 

Join (7^ and DH. 
Upon Q'D' (= CD) construct the A C'WD'^A CHD, 

and draw A^H\ 
Now ABCH > A'B'C^W, § 447 

and AEDH > A'E'UW, 

{of aU j>olygon8 with sides all given hut one, the mcunmum can he inscrihed 
in a semicircle having the undetermined side for its diameter). 

Add these two inequalities, then 

ABCHDE> A'BCPH'UE\ 

Take away from the two figures the equal A QHD and Q^BPD\ 

Then ABODE > A'BC'D'EK q. e. ^ 



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234 



PLANE GEOMETEY. — BOOK V. 



Peoposition XXII. Theorem. 

449. Of isoperimetric polygons of the same nurriber 
of sides f the iruucimum is equilateral. 

B K 





Let ABCD etc., be the maximum of isoperimetric 
polygons of any given number of sides. 

To prove AB, BC, CDy etc., equal. 

Proof. Draw AC. 

The A ABC mvAi be the maximum of all the A which are 
formed upon ^(7 with a perimeter equal to that of A ABC. 

Otherwise, a greater A AKC could be substituted for 
A ABC, without changing the perimeter of the polygon. 

But this is inconsistent with the hypothesis that the poly- 
gon ABCD etc., is the maximum polygon. 

.*. the A ABC'i^ isosceles, § 446 

{of all ^ having the same base and equal perimeters^ the isosceles A is the 
maximum). 

In like manner it may be proved that BC= CD, etc. Q.E.a 
460. Cor. The maximum of isopeii^metric polygons of the 
same number of sides is a regular polygon. 

For, it is equilateral, § 449 

{the maxifnum of isoperim^Hc polygons of the same number of sides is 
equilateral). 

Also it can be inscribed in a circle, § 448 

{the maadmum of all polygons formed of given sides can be inscribed in a G). 
That is, it is equilateral and equiangular, 

and therefore regular. . § 395 

Q.E.O. 



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MAXIMA AND MINIMA. 



235 



Proposition XXIII. Theorem. 

461. Of isopeHTnetric regular polygons, that which 
ha& the greatest number of sides is the Tnajcimunv, 
o 





Let Q be a reffuiar polygon of three sides, and Qf 
a regular polygon of four sides, and let the two poly- 
gons have equal perimeters. 

To prove Q greater than Q, 

Proof. Draw CD from Cto any point in AB. 

Invert the A CD A and place it in the position DCE, let- 
ting D fall at (7, Cat D, and A at E. 

The polygon DBCE is an irregular polygon of four sides, 
which by construction has the same perimeter as Q\ and the 
same area as Q. 

Then the irregular polygon DBCE of four sides is less than 
the regular isoperimetric polygon Q of four sides. § 450 

In like manner it may be shown that Q' is less than a regular 
isoperimetric polygon of five sides, and so on. q. e. d. 

452. CoR. The area of a circle is greater than the area of 
any polygon of equal perimeter, 

T 382. Of all equivalent parallelograms having equal bases, the rec- 
tangle has the least perimeter. 

, 383. Of all rectangles of a given area, the square has the least 
perimeter. 

384. Of all triangles upon the same base, and having the same alti- 
tude, the isosceles has the least perimeter. 

"^385. To divide a straight line into two parts such that their product 
shall be a maximum. 



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236 



PLAKB GEOMETBY.— BOOK V. 



Peoposition XXrV. Theobem. 

468. Of regular polygons having a given area, tTuxt 
which has the greai>esb numiber of sides has the least 
perimeter. 



Q' 





Let Q B,nd Q' he regular polygons having the same 
area, and let Q' have the greater number of sides. 

To prove the perimeter of Q greater than the perimeter of Q\ 

Proof. Let Q" be a regular polygon having the same perim- 
eter as Q\ and the same number of sides as Q. 

Then Q!>Q!\ §451 

{of isoperimetric regular polygons, that which has the greatat number of 
sides is the maximum). 



But Q=Q'. 

/. the perimeter oi Q> the perimeter of Q" 
But the perimeter of Q' = the perimeter of Q", 
.'. the perimeter oi Q > that of Q'. 



of Q". ' 



Cons. 



Q.E.D. 



454. Cor. The circumference of a circle is less than the 
perimeter of any polygon of equal area. 



386. To inscribe in a semicircle a rectangle having a given area; 
a rectangle having the maximum area. 

387. To find a point in a ^mi-circumference such that the sum of its 
distances from the extremities of the diameter shia.ll be a maximum. 



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EXEROISIS. 237 



Theorems. 

388. The side of a circnmscribed equilateral triangle is equal to twice , 
the side of the similar inscribed triangle. Find the ratio of their areas. / 

389. The apothem of an inscribed equilateral triangle is equal to hajf 
the radius of the circle. 

390. The apothem of an inscribed regular hexagon is equal to half 
the side of the inscribed equilateral triangle. 

^91. The area of an inscribed regular hexagon is equal to three- 
fourths that of the circumscribed regular hexagon. 

392. The area of an inscribed regular hexagon is a mean proportional 
between the areas of the inscribed and the circumscribed equilateral 
triangles. 

393. The area of an inscribed regular octagon is equal to that of a 
rectangle whose sides are equal to the sides of the inscribed and the cir- 
cumscribed squares. 

^94. The area of an inscribed regular dodecagon is equal to three 
times the square of the radius. 

^ 395. Every equilateral polygon circumscribed about a circle is regu- 
lar if it has an odd number of sides. 

396. Every equiangular polygon inscribed in a circle is regular if it 
has an odd number of sides. 

397. Every equiangular polygon circumscribed about a circle is 
regular. 

398. Upon the six sides of a regular hexagon squares are constructed 
outwardly. Prove that the exterior vertices of these squares are the ver- 
tices of a regular dodecagon. 

»99. The alternate vertices of a regular hexagon are joined by straight 
lines. Prove that another regular hexagon is thereby formed. Find the 
ratio of the areas of the two hexagons. 

400. The radius of an inscribed regular polygon is the mean propor- 
tional between its apothem and the radius of the similar circumscribed 
regular polygon. 

401. The area of a circular ring is equal to that of a circle whose 
diameter is a chord of the outer circle and a tangent to the inner circle. 

402. The square of the side of an inscribed regular pentagon is equal 
to the sum of the squares of the radius of the circle and the side of the 
inscribed regular decagon. 






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238 PLANE GEOMETRY. — BOOK V. 

If B denotes the radius of a circle, and a one side of a regular inscribed 
polygon, show that : ' _^-- ; — > 

^ "' 403. In a regular pentagon, a « — /v^iO — 2V5. ^ 
404. In a regular octagon, a -= ^ V2— ■\/2. 



405. In a regular dodecagon, a = B V2— y/S, 

f 406. If on the legs of a right triangle, as diameters, semicircles are 
described external to the triangle, and from the whole figure a semicircle 
on the hypotenuse is Bubtracted, the remainder is equivalent to the given 
triangle. 

Numerical Exercises. 

407. The radius of a circle =r. Find one side of the circumscribed 
equilateral triangle. o! V "7 

408. The radius of a circle « r. Find one side of the circumscribed 
regular hexagon. ^A^^J 

,- 409. If the radius of a circle is r, and the side of an inscribed regular 
polygon is a, show that the side of the similar circumscribed regular 
polygon is equal to 2ar ^ 

V4r«-a« 

410. The radius of a circle = r. Prove that the area of the inscribed 
regular octagon is equal to 2r*'\/5.I-r 

411. The sides of three regular octagons are 3 feet, 4 feet, and 5 feet, 
respectively. Find the side of a regular octagon equal in area to the 
sum of the areas of the three given octagons. 

412. What is the width of the ring between two concentric circum- 
ferences whose lengths are 440 feet and 330 feet? 

/ 413. Find the angle subtended at the centre by an arc 6 feet 5 inches 
long, if the radius of the circle is 8 feet 2 inches. 

/ 414. Find the angle subtended at the centre of a circle by an arc 
whose length is equal to the radius of the circle. 

415. What is the length of the arc subtended by one side of a regular 
dodecagon inscribed in a circle whose radius is 14 feet? 

416. Find the side of a, square equivalent to a circle whose radius is 
56 feet. 



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EXERCISES. 239 

417. Find the area of a circle inscribed in a sqnare containing 196 
square feet- 

418. The diameter of a circular grass plot is 28 feet. Find the diam- 
eter of a circular plot just twice as large. 

419. Find the side of the largest square that can be cut out of a cir- 
cular piece of wood whose radius is 1 foot 8 inches. 

420. The radius of a circle is 3 feet. What is the radius of a circle 25 
times as large ? ^ as large ? -^ as large ? 

421. The radius of a circle is 9 feet. What are the radii of the con- 
centric circumferences that will divide the circle into three equivalent 
parts ? , 

^ 122. The chord of half an arc is 12 feet, and the radius of the circle is 
1 o feet. Find the height of the arc. 

423. The chord of an arc is 24 inches, and the height of the arc is 9 
inches. Find the diameter of the circle. 

424. Find the area of a sector, if the radius of the circle is 28 feet, 
and the angle at the centre 22^°. 

425. The radius of a circle = r. Find the area of the segment sub- 
tended by one side of the inscribed regular hexagon. 

426. Three equal circles are described, each touching the other two. 
If the common radius is r, find the area contained between the circles. 

Problems. 

To circumscribe about a given circle : 

427r An equilateral triangle. 429. A regular hexagon. 

428.. A square. 430. A regular octagon. 

431. To draw through a given point a line so that it shall divide a 
given circumference into two parts having the ratio 3 : 7. 
^ 132. To construct a circumference equal to the sum of two given 
(^umferences. 

433. To construct a circle equivalent to the sum of two given circles. 

434. To construct a circle equivalent to three times a given circle. 

435. To construct a circle equivalent to three-fourths of a given circle. 
To divide a given circle by a concentric circumference : 

436. Into two equivalent parts. 437. Into five equivalent parts. 



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240 PLANE GEOMETRY. — BOOK V. 

Miscellaneous Exercises. 

Theorems. 

438. The line joining the feet of the perpendiculars dropped from the 
extremities of the hase of an isosceles triangle to the opposite sides is 
parallel to the hase. 

439. If AD bisect the angle A of a triangle ABC, and BD bisect the 
exterior angle CBF, then angle ADB equals one-half angle ACB. 

440. The sum of the acute angles at the vertices of a pentagram (five- 
pointed star) is equal to^wo right angles. 

441. The bisectors of the angles of a parallelogram form a rectangle. 

442. The altitudes AD, BE, OF of the triangle ABC bisect the angles 
of the triangle DBF. 

Hint. Circles with AB, BC, AC as diameters will pass through F and 
2>, F &n^ F, D and F, respectively. 

443. f hfi portions of any straight line intercepted between the cir- 
cumferences of two concentric circles are equal. 

444. Two circles are tangent internally at P, and a chord AB of the 
larger circle touches the smaller circle at C. Prove that PC bisects the 
angle APB. 

Hint. Braw a common tangent at P, and apply J J 263, 269, 145. 

445. The diagonals of a trapezoid divide each other into segments 
which are proportional, 

446. The perpendiculars from two vertices of a triangle upon the 
opposite sides divide each other into segments reciprocally proportional. 

447. If through a point P in the circumference of a circle two chords 
are drawn, the chorda and the segments between P and a chord parallel 
to the tangent at P are reciprocally proportional. 

448. The perpendicular from any point of a circumference upon a 
chord is a mean proportional between the perpendiculars from the same 
point upon the tangents drawn at the extremities of the chord. 

449. In an isosceles right triangle either leg is a mean proportional 
between the hypotenuse and the perpendicular upon it from the vertex 
of the right angle. 

450. The area of a triangle is equal to half the product of its perim- 
eter by the radius of the inscribed circle. 



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MISCELLANEOUS EXERCISES. 241 

451. The perimeter of a triangle is to one side as the perpendicular 
from the opposite vertex is to the radius of the inscribed circle. 

452. The sum of the perpendiculars from any point within a convex 
equilateral polygon upon the sides is constant 

453. A diameter of a circle is divided into any two parts, and upon 
these parts as diameters semi-circumferences are described on opposite 
sides of the given diameter. Prove that the sum of their lengths is equal 
to the semi-circumference of the given circle, and that they divide the 
circle into two parts whose areas have the same ratio as the two parts 
into which the diameter is divided. 

454. Lines drawn from one vertex of a parallelogram to the middle 
points of the opposite sides trisect one of the diagonals. 

455. If two circles intersect in the points A and B^ and through A 
any secant CAD is drawn limited by the circumferences at C and D, the 
straight lines BQ BD, are to each other as the diameters of the circles. 

456. If three straight lines AA\ BB\ CC\ drawn from the vertices 
of a triangle ABC to the opposite sides, pass through a copimon point 
within the triangle, then 

AA^ BB^ ^ CCo^ 

457. Two diagonals of a regular pentagon, not drawn from a common 
Tertex, divide each other in extreme and mean ratio. 

Loci. 

458. Find the locns of a point P whose distances from two given 
points A and .8 are in a given ratio (m : n). 

459. OT is any straight line drawn from a fixed point O to the cir- 
cumference of a fixed circle ; in OP a point Q is taken such that OQ: OB 
is constant. Find the locus of Q. 

460. From a fixed point A a straight line AB is drawn to any point 
in a given straight line C!D, and then divided at P in a given ratio 
(m : n). Find the locus of the point P. 

461. Find the locus of a point whose distances from two given straight 
lines are in a given ratio. (The locus consists of two straight lines.) 

462. Find the locus of a point the sum of whose distances from two 
given straight lines is eqnal to a given length ib. (See Ex. 73.) 



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7. 



242 PLANE GEOMETRY. — BOOK V. 



Problems. 

463. Qiven the perimeters of a regular inscribed and a similar circnm- 
scribed polygon, to compute the perimeters of the regular inscribed and 
circumscribed polygons of double the number of sides. 

464. To draw a tangent to a given circle such that the segment inter- 
cepted between the point of contact and a given straight line shall have 
a given length. 

465. To draw a straight line equidistant from three given points. 

466. To inscribe a straight line of given length between two given 
circumferences and parallel to a given straight line. (See Ex. 137.) 

467. To draw through a given point a straight line so that its dis- 
tances from two other given points shall be in a given ratio (m : n). 

HnsTT. Divide the line joining the two other points in the given ratio. 

468. Construct a square equivalent to the sum of a given triangle 
and a given parallelogram. 

469. Construct a rectangle having the difference of its base and 
altitude equal to a given line, and its area equivalent to the sum of a 
given triangle and a given pentagon. 

470. Construct a pentagon similar to a given pentagon and equiva- 
lent to a given trapezoid. 

471. To find a point whose distances from three given straight lines 
shall be as the numbers m, n, and p. . (See Ex. 461.) 

472. Given two circles intersecting at the point A. To draw through 
A a secant BACBUch that AB shall he to AC in a given ratio (m : n). 

HiHT. Divide the line of centres in the given ratio. 

473. To construct a triangle, given its angles and its area. 

474. To construct an equilateral triangle having a given area. 

475. To divide a given triangle into two equal parts by a line drawn 
parallel to one of the sides. 

,*>476. Given three points A, B, C. To find a fourth point Psuch that 
the areas of the triangles APB, AFC, BJPC, shall be equal. 

477. To construct a triangle, given its base, the ratio of the other 
sides, and the angle included by them. 

478. To divide a given circle into any number of equivalent parts by 
concentric circumferences. 

479. In a given equilateral triangle, to inscribe three equal cirdes 
tangent to each other and to the sides of the triangle. 



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